Biostatistics Quiz: T Distribution

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T DistributionQuestion 1 of 20

A pharmaceutical company is conducting a dose-escalation study where they measure drug concentration in blood samples. Due to logistical constraints, they can only recruit 11 subjects for the initial phase.

The researchers calculate a 95% confidence interval for mean drug concentration as (12.3, 18.7) μg/mL. If they want to reduce the width of this interval by 25% while maintaining the same confidence level, and assuming the population standard deviation remains constant, approximately how many total subjects would they need?

Approximately 17 subjects, accounting primarily for the degrees of freedom effect on t-critical values

Approximately 15 subjects, since the relationship between sample size and interval width is linear

Approximately 19 subjects, since reducing interval width by 25% requires increasing sample size by approximately 78%

Approximately 22 subjects, since reducing interval width requires squaring the inverse reduction factor

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Biostatistics Quiz

Biostatistics Quiz: T Distribution

Practice T Distribution in Biostatistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on T Distribution, giving you a quick way to practice the rules, question types, and explanations that matter most for Biostatistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Approximately 17 subjects, accounting primarily for the degrees of freedom effect on t-critical values
Approximately 15 subjects, since the relationship between sample size and interval width is linear
Approximately 19 subjects, since reducing interval width by 25% requires increasing sample size by approximately 78% (correct answer)
Approximately 22 subjects, since reducing interval width requires squaring the inverse reduction factor

Explanation: When you encounter confidence interval problems involving changes to interval width, you're dealing with the fundamental relationship between sample size and precision. The key insight is that confidence interval width is inversely proportional to the square root of sample size. The width of a confidence interval is

2 \times t \times \frac{s}{\sqrt{n}}

. To reduce width by 25% means the new width should be 75% of the original, or multiplied by 0.75. Since width is inversely proportional to

\sqrt{n}

, if you want the width to be 0.75 times smaller, you need

\sqrt{n}

to be

\frac{1}{0.75} = 1.33

times larger. Therefore, the new sample size must be

(1.33)^2 = 1.78

times the original size. With 11 original subjects:

11 \times 1.78 ≈ 19

subjects. Choice A incorrectly focuses on the t-distribution's degrees of freedom effect, which is minimal and doesn't drive the primary relationship. Choice B falsely assumes a linear relationship between sample size and interval width – this is wrong because the relationship involves a square root. Choice D correctly identifies that you need to square something, but applies it incorrectly to the reduction factor itself rather than the reciprocal. Study tip: Remember the "inverse square root rule" for confidence intervals: to cut the width in half, you need four times the sample size. For any proportional width reduction, divide 1 by that proportion, then square the result to find the sample size multiplier.

Question 2

A biostatistician receives conflicting advice about analyzing data from 8 subjects. Advisor A recommends using df = 7 for the t-test because "you always use n-1 degrees of freedom." Advisor B suggests using df = 6 because "with such a small sample, you need to be more conservative." Advisor C proposes df = 8 because "the sample size itself determines degrees of freedom." Which advisor gives the most appropriate guidance for a standard one-sample t-test?

Advisor A is correct; degrees of freedom equal n-1 because one parameter (sample mean) is estimated from the data, reducing available information (correct answer)
Advisor B is correct; small samples require additional conservatism beyond the standard t-distribution to account for increased uncertainty
Advisor C is correct; degrees of freedom should equal sample size to fully utilize all available data points in the analysis
None are correct; degrees of freedom should be determined by the specific research design and number of treatment groups being compared

Explanation: In a one-sample t-test, degrees of freedom = n-1 because we estimate one parameter (the population mean) using the sample mean. This estimation 'uses up' one degree of freedom, leaving n-1 for the test statistic. This is a fundamental property of the t-distribution derivation, not a matter of conservatism or preference. Advisor B's approach would be overly conservative without theoretical justification, and Advisor C's approach would underestimate the uncertainty in the test.

Question 3

A sample of $n = 16$ observations from a normal distribution yields $\bar{x} = 52.4$ and $s = 8.2$ . The 95% confidence interval for the population mean uses $t_{0.025,15} = 2.131$ . What would happen to the width of this confidence interval if the sample size increased to $n = 25$ with the same sample mean and standard deviation?

The interval width would increase because the t-value decreases to $t_{0.025,24} = 2.064$
The interval width would decrease because both the t-value and standard error decrease (correct answer)
The interval width would remain the same because the confidence level stays at 95%
The interval width would increase because more observations increase variability
The interval width would decrease only because the standard error decreases from $\frac{8.2}{\sqrt{16}}$ to $\frac{8.2}{\sqrt{25}}$

Explanation: When you encounter confidence interval questions involving sample size changes, focus on how sample size affects both the t-value and the standard error in the confidence interval formula:

\bar{x} \pm t_{\alpha/2,df} \cdot \frac{s}{\sqrt{n}}

. As sample size increases from 16 to 25, two key changes occur that both work to narrow the interval width. First, the degrees of freedom increase from 15 to 24, causing the t-value to decrease from 2.131 to 2.064. Second, and more importantly, the standard error

\frac{s}{\sqrt{n}}

decreases significantly because you're dividing by

\sqrt{25} = 5

instead of

\sqrt{16} = 4

. With

s = 8.2

, the standard error drops from

\frac{8.2}{4} = 2.05

\frac{8.2}{5} = 1.64

. Both changes reduce the margin of error, making answer B correct. Answer A incorrectly suggests the interval widens despite acknowledging that the t-value decreases - this ignores the more substantial effect of the reduced standard error. Answer C reflects a fundamental misunderstanding: while the confidence level remains 95%, the interval width definitely changes because the underlying precision improves with more data. Answer D represents a common misconception that more observations increase variability, when actually larger samples provide more precise estimates of the population mean. Study tip: Remember that larger sample sizes always lead to narrower confidence intervals (assuming other factors remain constant) because

\frac{1}{\sqrt{n}}

decreases as

n

increases, and this effect typically outweighs the smaller decrease in t-values.

Question 4

A pharmaceutical company tests a new drug on $n = 9$ patients. The mean reduction in blood pressure is $\bar{x} = 12.6$ mmHg with standard deviation $s = 4.3$ mmHg. To test $H_0: \mu = 10$ versus $H_1: \mu > 10$ , what is the calculated t-statistic?

$t = \frac{12.6 - 10}{4.3} = 0.605$
$t = \frac{12.6 - 10}{\frac{4.3}{\sqrt{9}}} = 1.814$ (correct answer)
$t = \frac{12.6 - 10}{\frac{4.3}{\sqrt{8}}} = 1.713$
$t = \frac{12.6 - 10}{4.3^2} = 0.141$
$t = \frac{12.6 - 10}{\frac{4.3^2}{9}} = 1.265$

Explanation: When you encounter a one-sample t-test question, you're testing whether a sample mean differs significantly from a hypothesized population mean. The key is recognizing that you need the standard error of the mean, not just the sample standard deviation. The t-statistic formula for a one-sample test is:

t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}

, where

\bar{x}

is the sample mean,

\mu_0

is the hypothesized mean,

s

is the sample standard deviation, and

n

is the sample size. Here, you calculate:

t = \frac{12.6 - 10}{4.3/\sqrt{9}} = \frac{2.6}{4.3/3} = \frac{2.6}{1.433} = 1.814

. This matches answer choice B. Choice A incorrectly uses the sample standard deviation (4.3) directly in the denominator instead of the standard error. This is a common mistake that ignores sampling variability. Choice C uses

\sqrt{8}

instead of

\sqrt{9}

in the denominator. This error stems from confusing the sample size (n = 9) with degrees of freedom (df = n-1 = 8). While degrees of freedom matter for the critical value, the standard error calculation always uses the actual sample size. Choice D squares the standard deviation in the denominator, which has no statistical basis and dramatically inflates the denominator. Study tip: Remember the pattern: for one-sample t-tests, always divide the sample standard deviation by

\sqrt{n}

to get the standard error. The degrees of freedom (n-1) only come into play when finding critical values from the t-table, not when calculating the test statistic itself.

Question 5

As the degrees of freedom in a t-distribution increase from 5 to 30, which statement best describes what happens to the distribution?

The distribution becomes more skewed and the tails become thinner, approaching a chi-square distribution
The distribution remains symmetric but becomes more peaked, with thicker tails than the standard normal
The distribution becomes less peaked and approaches the standard normal distribution with thinner tails (correct answer)
The distribution shifts to the right and the variance increases due to additional degrees of freedom
The distribution becomes bimodal with peaks at ±1 and the mean shifts away from zero

Explanation: When analyzing how t-distributions behave as degrees of freedom change, you're examining one of the most important convergence properties in statistics. The t-distribution was specifically designed to handle situations with small sample sizes, and its shape depends entirely on the degrees of freedom parameter. As degrees of freedom increase from 5 to 30, the t-distribution undergoes a predictable transformation. With more degrees of freedom, the distribution becomes less peaked (lower kurtosis) and its tails become progressively thinner. Most importantly, it approaches the standard normal distribution as a limiting case. This happens because larger degrees of freedom indicate more information in your sample, reducing the uncertainty that creates the t-distribution's characteristic heavy tails. Answer C correctly captures this behavior - the distribution becomes less peaked and approaches normality with thinner tails. Answer A is wrong because the t-distribution always remains symmetric (never skewed) and has no relationship to the chi-square distribution. Answer B incorrectly states the distribution becomes more peaked with thicker tails - this is exactly backwards from what actually occurs. Answer D is wrong because t-distributions don't shift laterally (they remain centered at zero) and while variance does decrease with more degrees of freedom, the distribution doesn't shift right. Remember this key principle: as degrees of freedom increase, t-distributions always converge toward the standard normal. By about 30 degrees of freedom, they're practically indistinguishable from

N(0,1)

. This is why many statistical tables switch from t to z at around 30 df.

Question 6

A quality control engineer takes a random sample of $n = 7$ measurements from a normally distributed process. If $\bar{x} = 24.8$ and $s = 2.1$ , what is the 90% confidence interval for the population mean using $t_{0.05,6} = 1.943$ ?

$24.8 \pm 1.943 \times \frac{2.1}{\sqrt{7}} = 24.8 \pm 1.541$ or $(23.26, 26.34)$ (correct answer)
$24.8 \pm 1.943 \times \frac{2.1}{\sqrt{6}} = 24.8 \pm 1.665$ or $(23.14, 26.46)$
$24.8 \pm 1.943 \times 2.1 = 24.8 \pm 4.080$ or $(20.72, 28.88)$
$24.8 \pm 1.943 \times \frac{2.1^2}{7} = 24.8 \pm 1.225$ or $(23.58, 26.02)$
$24.8 \pm 1.943 \times \frac{2.1}{7} = 24.8 \pm 0.583$ or $(24.22, 25.38)$

Explanation: When you encounter a confidence interval problem with a small sample from a normal distribution, you need to use the t-distribution formula:

\bar{x} \pm t_{\alpha/2,df} \times \frac{s}{\sqrt{n}}

, where the degrees of freedom equal

n-1

. Let's work through this systematically. You have

n=7

measurements, so

df = 7-1 = 6

. For a 90% confidence interval, you need

t_{0.05,6} = 1.943

(given). The standard error of the mean is

\frac{s}{\sqrt{n}} = \frac{2.1}{\sqrt{7}} = 0.794

. Therefore:

24.8 \pm 1.943 \times 0.794 = 24.8 \pm 1.541

, giving you the interval

(23.26, 26.34)

. Answer A correctly applies this formula. Answer B makes a common error by using

\sqrt{6}

instead of

\sqrt{7}

in the denominator—perhaps confusing the degrees of freedom (6) with the sample size (7) in the standard error calculation. Answer C omits the square root entirely, using just

s

instead of

\frac{s}{\sqrt{n}}

, which dramatically overestimates the margin of error. Answer D incorrectly squares the standard deviation and divides by

n

without taking the square root, creating a completely wrong formula. Study tip: Always double-check that you're using

\sqrt{n}

(not

n

\sqrt{df}

) in the denominator of the standard error, and remember that degrees of freedom for a single sample is always

n-1

, not

n

Question 7

Two researchers independently estimate the mean of the same population using t-distributions. Researcher A uses $n_A = 10$ observations, while Researcher B uses $n_B = 20$ observations. Both construct 95% confidence intervals with identical sample means and standard deviations. How do their intervals compare?

Researcher A's interval is wider because $t_{0.025,9} = 2.262 > t_{0.025,19} = 2.093$ and both have the same standard error
Researcher B's interval is wider because larger sample sizes always produce wider intervals due to increased precision
Researcher A's interval is wider because both the t-value is larger and the standard error $\frac{s}{\sqrt{n}}$ is larger (correct answer)
The intervals have identical width because they use the same confidence level and sample statistics
Researcher B's interval is wider only because the standard error $\frac{s}{\sqrt{20}}$ is larger than $\frac{s}{\sqrt{10}}$

Explanation: When comparing confidence intervals constructed using t-distributions, you need to consider both components of the margin of error: the critical t-value and the standard error of the mean. The confidence interval formula is:

\bar{x} \pm t_{\alpha/2,df} \cdot \frac{s}{\sqrt{n}}

Let's examine how sample size affects each component. The critical t-value decreases as degrees of freedom increase:

t_{0.025,9} = 2.262

versus

t_{0.025,19} = 2.093

. So Researcher A has a larger t-value. More importantly, the standard error

\frac{s}{\sqrt{n}}

decreases as sample size increases. With identical standard deviations, Researcher A's standard error is

\frac{s}{\sqrt{10}}

while Researcher B's is

\frac{s}{\sqrt{20}} = \frac{s}{\sqrt{10}} \cdot \frac{1}{\sqrt{2}}

, making A's standard error larger by a factor of

\sqrt{2} \approx 1.41

. Answer C correctly identifies that Researcher A's interval is wider because both the t-value AND the standard error are larger. Answer A incorrectly claims both researchers have the same standard error—this ignores the sample size difference. Answer B has the relationship backwards; larger samples produce narrower intervals due to reduced variability, not wider ones. Answer D ignores that identical confidence levels don't guarantee identical widths when sample sizes differ. Study tip: Remember that confidence interval width depends on both the critical value AND the standard error. Larger samples always reduce interval width through both smaller t-values and smaller standard errors.

Question 8

A nutritionist measures vitamin C content in $n = 8$ orange samples. The data follows a normal distribution with $\bar{x} = 65.4$ mg and $s = 7.2$ mg. If she wants to test $H_0: \mu = 60$ mg versus $H_1: \mu \neq 60$ mg at $\alpha = 0.01$ , what critical value(s) should she use?

$\pm t_{0.005,7} = \pm 3.499$ because it's a two-tailed test with $\alpha = 0.01$ and $df = 7$ (correct answer)
$t_{0.01,7} = 2.998$ because the alternative hypothesis suggests a one-tailed test
$\pm t_{0.01,8} = \pm 2.896$ because the sample size is 8 and $\alpha = 0.01$
$\pm z_{0.005} = \pm 2.576$ because the sample follows a normal distribution
$\pm t_{0.005,8} = \pm 3.355$ because we need the 99% confidence level with $df = 8$

Explanation: When you encounter hypothesis testing with small samples and unknown population standard deviation, you need to determine the correct distribution and critical values. This scenario calls for a t-test since we have

n = 8

(small sample) and only know the sample standard deviation

s = 7.2

mg. The correct approach uses

\pm t_{0.005,7} = \pm 3.499

. Here's why: The alternative hypothesis

H_1: \mu \neq 60

indicates a two-tailed test, so you split

\alpha = 0.01

between both tails, giving

\alpha/2 = 0.005

in each tail. The degrees of freedom equal

n - 1 = 8 - 1 = 7

. Therefore, you need

t_{0.005,7} = 3.499

, and since it's two-tailed, the critical values are

\pm 3.499

. Option B incorrectly suggests this is a one-tailed test. The

\neq

symbol in

H_1

always indicates two tails, not one. Option C makes two errors: it uses the wrong degrees of freedom (8 instead of 7) and fails to split alpha properly for a two-tailed test, using

t_{0.01,8}

instead of

t_{0.005,7}

. Option D incorrectly applies the z-distribution. Even though the population is normally distributed, when

\sigma

is unknown and estimated by

s

, you must use the t-distribution regardless of the population's distribution. Study tip: Remember the decision tree: unknown

\sigma

= t-test, known

\sigma

= z-test. For two-tailed tests, always divide your alpha by 2 when finding critical values, and degrees of freedom for one-sample t-tests always equal

n - 1

Question 9

A psychologist compares reaction times before and after caffeine consumption in $n = 15$ subjects. The differences (after - before) have mean $\bar{d} = -12.4$ ms and standard deviation $s_d = 18.6$ ms. What type of t-test should be performed and what are the appropriate degrees of freedom?

Paired t-test with $df = 14$ because we have matched pairs from the same subjects (correct answer)
Two-sample t-test with $df = 28$ because we're comparing two different conditions
One-sample t-test with $df = 15$ because we have one sample of difference scores
Two-sample t-test with $df = 14$ because we lose one degree of freedom for each sample
Paired t-test with $df = 30$ because we have 15 before and 15 after measurements

Explanation: When analyzing experiments where the same subjects are measured under two different conditions, you need to identify whether you're dealing with independent groups or matched pairs. This distinction determines both your statistical test and degrees of freedom calculation. Answer A is correct because this is a classic paired t-test scenario. You have the same 15 subjects measured before and after caffeine consumption, creating matched pairs. The key indicator is that you're given difference scores (

\bar{d} = -12.4

ms and

s_d = 18.6

ms), which is exactly what paired t-tests analyze. With

n = 15

pairs, your degrees of freedom equal

n - 1 = 14

. Answer B is wrong because while you are comparing two conditions, they're not independent samples. The

df = 28

would only apply if you had 15 subjects in one group and 15 different subjects in another group. Here, the same people are measured twice, violating the independence assumption of two-sample tests. Answer C is wrong because although you do have one sample of difference scores, this misses the fundamental design. You don't simply have 15 random differences—you have 15 paired differences from a before-after design. Also,

df = 15

incorrectly suggests you don't lose any degrees of freedom. Answer D is wrong because it correctly identifies needing

df = 14

but incorrectly calls for a two-sample test, which doesn't apply to paired data. Study tip: Look for keywords like "same subjects," "before and after," or "matched pairs," and check if difference scores are provided—these signal paired t-tests with

df = n - 1

Question 10

A biologist measures enzyme activity in $n = 13$ samples and obtains $\bar{x} = 42.7$ units with $s = 6.4$ units. She tests $H_0: \mu = 45$ versus $H_1: \mu < 45$ and calculates $t = -1.297$ . Using $t_{0.10,12} = 1.356$ and $t_{0.05,12} = 1.782$ , what can she conclude at $\alpha = 0.10$ ?

Reject $H_0$ because $|t| = 1.297 < 1.356$ , providing sufficient evidence that $\mu < 45$
Fail to reject $H_0$ because $t = -1.297 > -1.356$ , so the test statistic doesn't fall in the rejection region (correct answer)
Reject $H_0$ because $t = -1.297 < 0$ , indicating the sample mean is less than the hypothesized value
Fail to reject $H_0$ because we need $|t| > 1.782$ for significance at any reasonable level
The test is inconclusive because $t = -1.297$ falls between the critical values $\pm 1.356$

Explanation: When you encounter a one-tailed t-test problem, focus on identifying the correct rejection region based on the alternative hypothesis direction and comparing your test statistic to the critical value. This is a left-tailed test since

H_1: \mu < 45

. For

\alpha = 0.10

with 12 degrees of freedom, the critical value is

-t_{0.10,12} = -1.356

. You reject

H_0

only if your test statistic falls in the rejection region, meaning

t < -1.356

. Since

t = -1.297 > -1.356

, the test statistic doesn't fall in the rejection region, so you fail to reject

H_0

. Answer B correctly identifies this logic: the test statistic

t = -1.297

is greater than the critical value

-1.356

, placing it outside the rejection region. Answer A makes a critical error by using the absolute value

|t| = 1.297 < 1.356

and incorrectly concluding to reject

H_0

. This treats it like a two-tailed test and misinterprets the comparison. Answer C shows a fundamental misunderstanding—just because

t < 0

doesn't automatically mean you reject

H_0

. The test statistic must be sufficiently negative (beyond the critical value) to provide strong enough evidence. Answer D incorrectly references

t_{0.05,12} = 1.782

, which isn't relevant since we're testing at

\alpha = 0.10

, not 0.05. Study tip: For one-tailed tests, always determine the correct critical value sign based on the alternative hypothesis direction, then check if your test statistic falls beyond that critical value in the tail of interest.

Question 11

A quality analyst compares the tensile strength of materials from two suppliers using samples of $n_1 = 7$ and $n_2 = 9$ . She assumes equal variances and calculates a pooled t-statistic of $t = 2.85$ . For testing equality of means at $\alpha = 0.01$ , what should she conclude if $t_{0.005,14} = 2.977$ ?

Reject $H_0$ because $t = 2.85 > t_{0.005,14} = 2.977$ , indicating a significant difference
Fail to reject $H_0$ because $t = 2.85 < t_{0.005,14} = 2.977$ , providing insufficient evidence (correct answer)
Reject $H_0$ because $t = 2.85 > 0$ , indicating the first supplier has higher strength
Fail to reject $H_0$ because the degrees of freedom should be $df = 15$ , not $df = 14$
The test is inconclusive because $t = 2.85$ is very close to the critical value

Explanation: When you encounter a two-sample t-test with pooled variances, you're testing whether two population means are significantly different. The key is properly comparing your calculated t-statistic to the critical value at your chosen significance level. Here's the setup: You have

n_1 = 7

and

n_2 = 9

samples, giving you

df = n_1 + n_2 - 2 = 14

degrees of freedom. For a two-tailed test at

\alpha = 0.01

, you need

t_{0.005,14} = 2.977

as your critical value (the 0.005 comes from splitting the 0.01 alpha between two tails). Since your calculated

t = 2.85

is less than the critical value of 2.977, you fail to reject the null hypothesis. The difference isn't statistically significant at the 1% level. Option A incorrectly states that 2.85 > 2.977, which is mathematically wrong. Option C makes the mistake of thinking any positive t-value indicates significance—but statistical significance depends on exceeding the critical value, not just being positive. Option D incorrectly calculates degrees of freedom; for two independent samples, it's always

n_1 + n_2 - 2

, not

n_1 + n_2 - 1

. Study tip: Always double-check your inequality direction when comparing test statistics to critical values. For rejection, your calculated statistic must exceed the critical value in absolute terms. Many students mix this up under exam pressure, so practice writing out "reject if

|t| > t_{critical}

" until it's automatic.

Question 12

Two independent samples are drawn from normal populations. Sample 1 has $n_1 = 8$ observations, and Sample 2 has $n_2 = 15$ observations. Both population variances are unknown but assumed equal. What degrees of freedom should be used for the pooled t-test?

$df = 21$ because $n_1 + n_2 - 2 = 8 + 15 - 2$ (correct answer)
$df = 23$ because $n_1 + n_2 = 8 + 15$
$df = 7$ because $\min(n_1 - 1, n_2 - 1) = \min(7, 14)$
$df = 14$ because $\max(n_1 - 1, n_2 - 1) = \max(7, 14)$
$df = 11$ because $\frac{(n_1 - 1) + (n_2 - 1)}{2} = \frac{7 + 14}{2}$

Explanation: When you encounter a two-sample t-test with unknown but equal variances, you're dealing with a pooled t-test scenario. The key insight is that you're combining information from both samples to estimate the common population variance, which affects how many degrees of freedom remain for the test. The pooled t-test uses both sample variances to create a single, more reliable estimate of the shared population variance. Since you're estimating one mean from each sample (using up one degree of freedom per sample), the total degrees of freedom equals the combined sample sizes minus the number of means estimated:

df = n_1 + n_2 - 2

. With

n_1 = 8

and

n_2 = 15

, this gives us

df = 8 + 15 - 2 = 21

. Answer A correctly applies this formula. Answer B (

df = 23

) makes the common mistake of forgetting to subtract the 2 degrees of freedom lost when estimating the sample means. Answer C (

df = 7

) incorrectly uses only the smaller sample's degrees of freedom, which would apply to certain non-parametric tests but not the pooled t-test. Answer D (

df = 14

) similarly uses only the larger sample's degrees of freedom, ignoring the contribution of the smaller sample entirely. Study tip: Remember the pattern "total observations minus parameters estimated." For pooled t-tests, you always estimate exactly 2 parameters (the two sample means), so the formula is always

n_1 + n_2 - 2

, regardless of which sample is larger.

Question 13

A researcher wants to test whether the mean weight loss from a diet program differs from 5 pounds. With $n = 14$ participants, she calculates a t-statistic of $t = -2.45$ . Using $\alpha = 0.05$ for a two-tailed test, what can she conclude if $t_{0.025,13} = 2.160$ ?

Reject $H_0$ because $|-2.45| = 2.45 > 2.160$ , so the mean differs significantly from 5 pounds (correct answer)
Fail to reject $H_0$ because $t = -2.45 < 0$ , indicating insufficient evidence against the null hypothesis
Reject $H_0$ because $t = -2.45 < -2.160$ , providing strong evidence that mean weight loss exceeds 5 pounds
Fail to reject $H_0$ because the test statistic $t = -2.45$ falls within the acceptance region $(-2.160, 2.160)$
The test is inconclusive because the degrees of freedom should be $df = 14$ rather than $df = 13$

Explanation: When you encounter a two-tailed t-test question, focus on whether the absolute value of your test statistic exceeds the critical value. This determines if your result falls in the rejection region. In this weight loss study, you're testing

H_0: \mu = 5

versus

H_1: \mu \neq 5

pounds. With a two-tailed test at

\alpha = 0.05

, you reject the null hypothesis if

|t| > t_{0.025,13} = 2.160

. Since

|-2.45| = 2.45 > 2.160

, you reject

H_0

and conclude the mean weight loss differs significantly from 5 pounds. Looking at the wrong answers: Option B incorrectly assumes that a negative t-statistic alone indicates insufficient evidence. The sign of t only tells you the direction of the difference, not its significance. Option C makes two errors—it uses the wrong comparison (

t < -2.160

instead of

|t| > 2.160

) and incorrectly interprets what rejecting

H_0

means about the direction of weight loss. Option D incorrectly describes the acceptance region as

(-2.160, 2.160)

, but since

|-2.45| = 2.45

exceeds 2.160, the test statistic actually falls in the rejection region. Study tip: For two-tailed tests, always work with absolute values when comparing to critical values. The rejection region consists of both tails:

t < -t_{\alpha/2}

t > t_{\alpha/2}

, which simplifies to

|t| > t_{\alpha/2}

. Don't let negative test statistics confuse you—focus on the magnitude.

Question 14

Two independent samples from normal populations have $n_1 = 6$ and $n_2 = 10$ observations. If the population variances are unknown and unequal, what is the approximate degrees of freedom for Welch's t-test using $s_1^2 = 16$ and $s_2^2 = 25$ ?

$df = \frac{(\frac{16}{6} + \frac{25}{10})^2}{\frac{(16/6)^2}{5} + \frac{(25/10)^2}{9}} = \frac{(5.167)^2}{1.421 + 0.694} = 12.7$ (correct answer)
$df = \frac{(\frac{16}{6} + \frac{25}{10})^2}{\frac{(16/6)^2}{6} + \frac{(25/10)^2}{10}} = \frac{(5.167)^2}{1.184 + 0.625} = 14.7$
$df = n_1 + n_2 - 2 = 6 + 10 - 2 = 14$ because this is a two-sample test
$df = \min(n_1-1, n_2-1) = \min(5, 9) = 5$ for the conservative approach
$df = \frac{s_1^2 + s_2^2}{\frac{s_1^2}{n_1-1} + \frac{s_2^2}{n_2-1}} = \frac{16 + 25}{\frac{16}{5} + \frac{25}{9}} = 8.9$

Explanation: When you encounter a two-sample t-test with unequal variances, you need Welch's t-test, which uses a special formula to approximate the degrees of freedom. This situation arises when you can't assume equal population variances, making the standard pooled t-test inappropriate. The correct approach uses the Welch-Satterthwaite equation:

df = \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}}

Substituting our values:

df = \frac{(\frac{16}{6} + \frac{25}{10})^2}{\frac{(16/6)^2}{5} + \frac{(25/10)^2}{9}} = \frac{(2.667 + 2.5)^2}{\frac{7.111}{5} + \frac{6.25}{9}} = \frac{26.69}{1.422 + 0.694} = 12.6

This matches option A's calculation and result of 12.7. Option B incorrectly uses

n_1

and

n_2

in the denominator instead of

(n_1-1)

and

(n_2-1)

, which inflates the degrees of freedom. Option C gives you the degrees of freedom for a pooled t-test assuming equal variances—this would be wrong when variances are clearly unequal. Option D represents an overly conservative approach that some statisticians might use, but it's not the standard Welch approximation. Study tip: Always check whether the problem states "equal variances" or "unequal variances." If unequal (or unknown and you suspect they're different), use Welch's formula with

(n-1)

terms in the denominator, not the sample sizes themselves.

Question 15

A pharmacist tests the dissolution time of $n = 9$ tablets and finds $\bar{x} = 18.6$ minutes with $s = 3.2$ minutes. She constructs a 95% confidence interval and gets $(16.19, 21.01)$ . If one additional tablet is tested and shows a dissolution time that makes the new sample mean $\bar{x}_{new} = 19.1$ minutes, how will this affect the confidence interval width?

The width will decrease because the larger sample size reduces both the t-value and the standard error (correct answer)
The width will increase because adding data increases the sample standard deviation significantly
The width will remain approximately the same because one additional observation has minimal impact
The width will increase because the new mean is farther from the original confidence interval center
The width will decrease only if the new standard deviation remains close to 3.2 minutes

Explanation: When you encounter confidence interval questions involving sample size changes, focus on how the width formula responds:

\text{Width} = 2 \times t_{\alpha/2} \times \frac{s}{\sqrt{n}}

Adding one tablet increases the sample size from

n = 9

n = 10

. This creates two beneficial effects that both reduce the interval width. First, the degrees of freedom increase from 8 to 9, which decreases the critical t-value (from approximately 2.306 to 2.262 for 95% confidence). Second, and more importantly, the standard error

\frac{s}{\sqrt{n}}

decreases because you're dividing by

\sqrt{10}

instead of

\sqrt{9}

. Even if the sample standard deviation

s

changes slightly, the

\sqrt{n}

effect dominates. Answer A correctly identifies that both the t-value and standard error decrease with larger sample size, narrowing the interval. Answer B incorrectly assumes the standard deviation will increase significantly—one additional data point rarely causes dramatic changes in

s

, and even if it did, the

\sqrt{n}

increase would likely offset this. Answer C misses that going from

n = 9

n = 10

represents an 11% sample size increase, which produces noticeable effects on confidence interval width. Answer D confuses confidence interval center movement with width changes—the new mean of 19.1 is well within the original interval (16.19, 21.01), and interval width depends on variability measures, not mean location. Remember: Larger sample sizes almost always produce narrower confidence intervals due to reduced standard error, regardless of minor changes in the sample mean or standard deviation.

Question 16

A researcher uses a t-distribution with $df = 24$ for hypothesis testing. If she increases her sample size so that $df = 49$ , approximately how much will the critical value $t_{0.025}$ change for a two-tailed test at $\alpha = 0.05$ ?

The critical value will decrease from approximately 2.064 to 2.010, a reduction of about 0.054 (correct answer)
The critical value will increase from approximately 2.064 to 2.120, an increase of about 0.056
The critical value will decrease from approximately 2.064 to 1.960, approaching the z-value
The critical value will remain approximately 2.064 because both are large sample sizes
The critical value will decrease from approximately 2.064 to 1.677, a substantial reduction

Explanation: When you encounter t-distribution problems involving degrees of freedom changes, remember that as sample size increases (and thus df increases), the t-distribution approaches the standard normal distribution, making critical values smaller in absolute value. Let's trace through this systematically. With

df = 24

, the critical value

t_{0.025}

for a two-tailed test at

\alpha = 0.05

is approximately 2.064. When the sample size increases so that

df = 49

, the critical value becomes approximately 2.010. This represents a decrease of about 0.054, confirming that answer A is correct. Looking at the wrong answers: Answer B incorrectly suggests the critical value increases to 2.120. This contradicts the fundamental property that t-values decrease as degrees of freedom increase. Answer C states the critical value drops all the way to 1.960 (the z-value), but this is premature—while the t-distribution does approach the normal distribution, it doesn't reach

z_{0.025} = 1.960

until much larger sample sizes. Answer D claims the critical value remains unchanged because both are "large sample sizes," but this ignores the meaningful difference between 25 and 50 observations. Study tip: Remember the key relationship: as df increases, t-critical values decrease toward their corresponding z-values. The convergence is gradual—by df = 30, you're close to z-values, but not identical. This pattern appears frequently on biostatistics exams, so practice recognizing when sample size changes affect your critical values.

Question 17

A medical researcher tests whether a new treatment reduces recovery time. She collects data from $n = 12$ patients and finds $\bar{x} = 8.2$ days with $s = 2.5$ days. For testing $H_0: \mu \geq 10$ versus $H_1: \mu < 10$ , what is the p-value if the calculated t-statistic is $t = -2.49$ ?

$P(T_{11} \leq -2.49)$ , which is the area to the left of -2.49 in the t-distribution with 11 df (correct answer)
$P(T_{11} \geq 2.49)$ , which is the area to the right of 2.49 in the t-distribution with 11 df
$2 \times P(T_{11} \leq -2.49)$ , which accounts for both tails of the distribution
$P(T_{12} \leq -2.49)$ , which uses the sample size as degrees of freedom
$P(T_{11} \geq -2.49)$ , which is the area to the right of -2.49 in the t-distribution with 11 df

Explanation: When you encounter a hypothesis test with a calculated t-statistic, determining the p-value requires understanding both the alternative hypothesis direction and the sampling distribution properties. This is a one-tailed (left-tailed) test because

H_1: \mu < 10

specifies a direction. The researcher wants to know if the treatment reduces recovery time below 10 days. With

n = 12

patients, you have

df = n - 1 = 11

degrees of freedom for the t-distribution. The p-value represents the probability of observing a test statistic as extreme or more extreme than what was calculated, assuming

H_0

is true. Since this is a left-tailed test and

t = -2.49

, you need the probability of getting a t-statistic of -2.49 or smaller (more negative). This is exactly

P(T_{11} \leq -2.49)

, making A correct. B is wrong because it uses the right tail, which would be appropriate for a right-tailed test (

H_1: \mu > 10

). C is incorrect because multiplying by 2 creates a two-tailed p-value, but this is a one-tailed test with a directional alternative hypothesis. D makes a degrees-of-freedom error by using the sample size (12) instead of

n-1 = 11

. Study tip: Always match the p-value calculation to your alternative hypothesis direction. Left-tailed tests use the left tail area, right-tailed tests use the right tail, and two-tailed tests double the smaller tail area. Remember that degrees of freedom for a one-sample t-test is always

n-1

Question 18

A researcher collects a sample of size $n = 12$ from a normally distributed population with unknown variance. If the sample mean is $\bar{x} = 45.2$ and the sample standard deviation is $s = 6.8$ , what is the appropriate distribution to use for constructing a confidence interval for the population mean?

Normal distribution with standard deviation $\sigma = 6.8$
t-distribution with $df = 11$ degrees of freedom (correct answer)
t-distribution with $df = 12$ degrees of freedom
Normal distribution with standard error $\frac{6.8}{\sqrt{12}}$
Chi-square distribution with $df = 11$ degrees of freedom

Explanation: When you encounter a confidence interval problem involving a sample from a normally distributed population, the key decision is whether to use a normal distribution or t-distribution. This choice depends on whether the population variance is known and the sample size. Since you have a small sample (

n = 12

) from a normally distributed population with unknown variance, you must use the t-distribution. The sample standard deviation

s = 6.8

is your estimate of the unknown population standard deviation, which introduces additional uncertainty that the t-distribution accounts for through its heavier tails. The degrees of freedom for a one-sample t-test or confidence interval is always

df = n - 1

. With

n = 12

, you get

df = 12 - 1 = 11

degrees of freedom, making answer B correct. Let's examine why the other options fail: Answer A incorrectly treats the sample standard deviation as if it were the known population standard deviation and uses the normal distribution. Answer C uses the t-distribution correctly but miscalculates the degrees of freedom as

n = 12

instead of

n - 1 = 11

. Answer D makes the same fundamental error as A by using the normal distribution when the population variance is unknown, even though it correctly calculates the standard error. Study tip: Remember the decision tree: unknown population variance + small sample (typically

n < 30

) from normal population = t-distribution with

df = n - 1

. Only use the normal distribution when the population variance is actually known, not estimated.

Question 19

A food scientist compares the protein content of two brands of cereal using independent samples. Brand A: $n_1 = 8$ , $\bar{x}_1 = 12.4$ g, $s_1 = 2.1$ g. Brand B: $n_2 = 12$ , $\bar{x}_2 = 10.8$ g, $s_2 = 1.9$ g. Assuming equal population variances, what is the pooled standard deviation?

$s_p = \sqrt{\frac{(8-1)(2.1)^2 + (12-1)(1.9)^2}{8+12-2}} = \sqrt{\frac{30.87 + 39.71}{18}} = 1.98$ (correct answer)
$s_p = \sqrt{\frac{8(2.1)^2 + 12(1.9)^2}{8+12}} = \sqrt{\frac{35.28 + 43.32}{20}} = 1.98$
$s_p = \frac{(8)(2.1) + (12)(1.9)}{8+12} = \frac{16.8 + 22.8}{20} = 1.98$
$s_p = \sqrt{\frac{(2.1)^2 + (1.9)^2}{2}} = \sqrt{\frac{4.41 + 3.61}{2}} = 2.00$
$s_p = \sqrt{(2.1)^2 + (1.9)^2} = \sqrt{4.41 + 3.61} = 2.83$

Explanation: When comparing means from two independent samples with assumed equal variances, you need the pooled standard deviation to estimate the common population standard deviation. This combines information from both samples to get a more reliable estimate than using either sample alone. The pooled standard deviation formula is

s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}}

. Notice that it weights each sample's variance by its degrees of freedom (

n-1

), not just the sample size. This gives you an unbiased estimate of the population variance. Choice A correctly applies this formula:

s_p = \sqrt{\frac{(8-1)(2.1)^2 + (12-1)(1.9)^2}{8+12-2}} = \sqrt{\frac{7(4.41) + 11(3.61)}{18}} = \sqrt{\frac{30.87 + 39.71}{18}} = 1.98

. Choice B incorrectly weights by sample sizes (8 and 12) rather than degrees of freedom (7 and 11), which would bias the estimate. Choice C makes two errors: it weights by sample sizes AND averages the standard deviations directly instead of variances, completely ignoring that variances (not standard deviations) are additive. Choice D simply averages the two variances without any weighting, throwing away valuable information about sample sizes. Remember: pooled variance calculations always use degrees of freedom as weights, and you must work with variances (

s^2

) in the calculation, then take the square root at the end to get back to standard deviation units.

Question 20

Two research teams study the same drug using identical protocols but different sample sizes. Team A (n=9) reports a 95% CI of (14.2, 21.8) mg/dL, while Team B (n=25) reports (15.1, 20.9) mg/dL. Assuming both teams observed the same sample mean and population standard deviation, what can be concluded about their analyses?

Both analyses appear correct, but Team A's interval is more reliable due to the more conservative t-distribution with fewer degrees of freedom
The analyses are inconsistent; Team A's wider interval suggests they may have used an inappropriate significance level or critical value
Both analyses are consistent; Team B's narrower interval reflects the improved precision from larger sample size and smaller t-critical value (correct answer)
The analyses are inconsistent; identical sample means and standard deviations should produce intervals with the same midpoint regardless of sample size

Explanation: When comparing confidence intervals from studies with different sample sizes, you need to understand how sample size affects interval width through the standard error and t-distribution critical values. Both teams observed the same sample mean, so their intervals should have identical midpoints. Team A's interval (14.2, 21.8) has a midpoint of 18.0, and Team B's interval (15.1, 20.9) also centers at 18.0, confirming consistent sample means. The key difference is interval width. Team B's larger sample size (n=25 vs n=9) provides two advantages: the standard error decreases by

\sqrt{n}

, and the t-critical value becomes smaller due to more degrees of freedom (24 vs 8). Both factors make Team B's interval narrower, reflecting improved precision from the larger sample. Answer C correctly identifies this relationship - both analyses are consistent, with Team B's narrower interval reflecting improved precision from larger sample size and smaller t-critical value. Answer A incorrectly suggests Team A's wider interval is "more reliable" and "conservative." While Team A does use a larger t-critical value due to fewer degrees of freedom, this doesn't make it more reliable - it's simply less precise. Answer B wrongly assumes the analyses are inconsistent and suggests Team A used inappropriate values. The wider interval is actually expected given the smaller sample size. Answer D correctly notes that identical means should produce the same midpoint, but incorrectly claims this makes the analyses inconsistent. Different interval widths are expected with different sample sizes. Study tip: Remember that larger sample sizes always produce narrower confidence intervals due to reduced standard error and smaller critical values - this indicates better precision, not inconsistency.