College Statistics Quiz: T Distribution And Degrees Of Freedom

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T Distribution And Degrees Of FreedomQuestion 1 of 20

A statistics student is creating a confidence interval for a population mean based on a random sample of size $n=7$ from a normal population with an unknown standard deviation. What is the variance of the t-distribution used to find the critical value?

1.50

1.25

1.20

1.00

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College Statistics Quiz

College Statistics Quiz: T Distribution And Degrees Of Freedom

Practice T Distribution And Degrees Of Freedom in College Statistics 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 And Degrees Of Freedom, giving you a quick way to practice the rules, question types, and explanations that matter most for College Statistics.

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

1.50 (correct answer)
1.25
1.20
1.00

Explanation: The t-distribution used for inference on a single mean has degrees of freedom

df = n-1

. In this case,

df = 7-1 = 6

. The theoretical variance of a t-distribution with

v

degrees of freedom is given by the formula

\text{Var}(T) = \frac{v}{v-2}

, which is defined for

v > 2

. Plugging in

v=6

, the variance is

\frac{6}{6-2} = \frac{6}{4} = 1.50

Question 2

Two independent studies (Study 1 and Study 2) test the same null hypothesis $H_0: \mu = 100$ against the alternative $H_A: \mu \neq 100$ at a significance level of $\alpha=0.01$ . Study 1 uses a sample of size $n_1 = 15$ , and Study 2 uses a sample of size $n_2 = 25$ . Which of the following statements is necessarily true regarding the two tests?

The critical value used to determine rejection will be larger for Study 1 than for Study 2.
If the null hypothesis is false, the probability of a Type II error will be greater for Study 1 than for Study 2.
The test statistic calculated for Study 1 must be smaller than the test statistic for Study 2.
Both A and B are necessarily true. (correct answer)

Explanation: Statement A is true: The critical value from a t-distribution depends on the degrees of freedom. Study 1 has

df_1 = 14

and Study 2 has

df_2 = 24

. For a fixed significance level, a t-distribution with fewer degrees of freedom has heavier tails, requiring a larger critical value to cut off the same tail area. Statement B is true: The probability of a Type II error (

\beta

) is inversely related to statistical power. Power increases with sample size. Since Study 2 has a larger sample size, it has more power to detect a true effect, and therefore a lower probability of a Type II error compared to Study 1. Since both statements A and B are necessarily true, D is the correct answer.

Question 3

The 95th percentile of a t-distribution with $v$ degrees of freedom is the value $t$ such that $P(T_v \leq t) = 0.95$ . How does the 95th percentile of a t-distribution with 5 degrees of freedom compare to the 95th percentile of a t-distribution with 50 degrees of freedom?

The 95th percentile is larger for $df=5$ than for $df=50$ . (correct answer)
The 95th percentile is smaller for $df=5$ than for $df=50$ .
The 95th percentiles are identical for both distributions.
The relationship depends on whether it's the 95th or the 5th percentile.

Explanation: The t-distribution with fewer degrees of freedom has heavier tails. This means that to capture a certain cumulative probability (like 95%), one must go further out into the tail of the distribution. For

P(T_v \leq t) = 0.95

, the value

t

must be larger for smaller

v

to encompass the extra probability mass in the right tail. For example, the 95th percentile for

df=5

is approximately 2.015, while for

df=50

it is approximately 1.676. Thus, the 95th percentile is larger for

df=5

Question 4

A researcher investigates the effect of a new public speaking workshop on students' anxiety levels. Anxiety is measured for 25 students before they take the workshop and again after they complete it. To test if the workshop significantly reduces anxiety, the researcher analyzes the differences in pre- and post-workshop scores. What are the degrees of freedom for the appropriate t-test?

24 (correct answer)
25
48
49

Explanation: This experimental design involves taking two measurements on the same set of subjects, which constitutes paired data. The correct statistical procedure is a paired-samples t-test. This test is performed by first calculating the difference between the pre- and post-workshop scores for each student, and then performing a one-sample t-test on these differences. The degrees of freedom for a one-sample t-test are

n-1

, where

n

is the number of pairs. Here, there are 25 students (pairs), so the degrees of freedom are

25 - 1 = 24

Question 5

An analyst is running a complex financial model where an intermediate parameter is known to follow a t-distribution. The analyst observes that Monte Carlo simulations of the model's variance are highly unstable and fail to converge. After investigation, it is determined that the parameter follows a t-distribution with 2 degrees of freedom. What is the statistical explanation for the simulation instability?

The mean of a t-distribution with 2 degrees of freedom is undefined, causing instability.
The variance of a t-distribution with 2 degrees of freedom is undefined. (correct answer)
Any simulation involving a t-distribution with fewer than 30 degrees of freedom is inherently unstable.
The t-distribution cannot be used in financial models, so the initial assumption was incorrect.

Explanation: The moments of a t-distribution depend on its degrees of freedom (

v

). The mean is defined for

v>1

and the variance is defined for

v>2

. The formula for the variance is

v/(v-2)

. For a t-distribution with exactly 2 degrees of freedom, the denominator of the variance formula is zero, meaning the theoretical variance is infinite or undefined. A Monte Carlo simulation attempting to estimate a quantity that is theoretically undefined will not converge to a stable value, which explains the observed instability.

Question 6

For a two-sided hypothesis test, a researcher calculates a test statistic of $t = 2.65$ based on a sample of size $n=20$ . A statistical software excerpt shows the following critical values for a t-distribution with 19 degrees of freedom: $t_{0.025} = 2.093$ , $t_{0.01} = 2.539$ , and $t_{0.005} = 2.861$ . Based on this information, which of the following is the tightest bound for the p-value?

$p < 0.01$
$0.01 < p < 0.02$ (correct answer)
$0.02 < p < 0.05$
$p > 0.05$

Explanation: The degrees of freedom are

df = n-1 = 20-1=19

. The calculated test statistic is

t=2.65

. We compare this value to the provided critical values for a t-distribution with 19 df. The value 2.65 lies between

t_{0.01} = 2.539

and

t_{0.005} = 2.861

. This means the probability in a single tail (the area to the right of 2.65) is between 0.005 and 0.01. Since the hypothesis test is two-sided, the p-value is twice the single-tail probability. Therefore, the p-value is between

2 \times 0.005

and

2 \times 0.01

, which gives the interval

0.01 < p < 0.02

. This is the tightest bound that can be determined from the given information.

Question 7

A biologist collects a small random sample of 15 fish from a lake to estimate the mean mercury level. The distribution of mercury levels in the fish population is unknown. Which of the following findings would provide the strongest justification for using a t-distribution to construct a confidence interval for the mean mercury level?

The sample standard deviation is very small relative to the sample mean, indicating low variability.
A histogram of the sample data shows a symmetric, unimodal shape with no obvious outliers. (correct answer)
The population of fish in the lake is known to be very large, ensuring sample independence.
Another study of a similar lake found that mercury levels there followed a normal distribution.

Explanation: The primary assumption for using the t-procedure with a small sample is that the data comes from a population that is approximately normally distributed. When the population distribution is unknown, the best available evidence for its shape comes from the sample itself. If a plot of the sample data (like a histogram or normal probability plot) appears symmetric, unimodal, and free of strong skew or outliers, it provides reasonable justification that the normality assumption is not severely violated. This is the strongest evidence among the choices for proceeding with the t-procedure.

Question 8

A researcher compares the probability of observing an extreme result under three different distributions: a standard normal distribution (Z), a t-distribution with 5 degrees of freedom ( $T_5$ ), and a t-distribution with 25 degrees of freedom ( $T_{25}$ ). Which of the following correctly orders the probabilities of observing a value greater than 3?

$P(Z > 3) < P(T_{25} > 3) < P(T_5 > 3)$ (correct answer)
$P(T_5 > 3) < P(T_{25} > 3) < P(Z > 3)$
$P(Z > 3) < P(T_5 > 3) < P(T_{25} > 3)$
The probabilities are approximately equal for such an extreme value and cannot be definitively ordered without a table.

Explanation: The t-distribution has heavier (or 'fatter') tails than the standard normal distribution. This means there is more probability in the tails. The heaviness of the tails is determined by the degrees of freedom (df); the lower the df, the heavier the tails. As df approaches infinity, the t-distribution converges to the standard normal distribution. Therefore, the t-distribution with the fewest degrees of freedom (

T_5

) will have the most probability in its tail beyond 3. The standard normal distribution (Z) will have the least. The

T_{25}

distribution will be in between. This gives the correct order:

P(Z > 3) < P(T_{25} > 3) < P(T_5 > 3)

Question 9

An experiment is conducted to compare the effectiveness of two different fertilizers on crop yield. 15 plots are treated with Fertilizer A, and 12 different plots are treated with Fertilizer B. The researchers assume the population variances of crop yields for the two fertilizers are not equal. To construct a confidence interval for the difference in mean yields, they use a two-sample t-procedure. Which of the following is the most appropriate and commonly used conservative value for the degrees of freedom?

11 (correct answer)
14
25
26

Explanation: When comparing two independent samples with unequal variances, Welch's t-test is used. The degrees of freedom for Welch's test are calculated with a complex formula (the Welch-Satterthwaite equation). However, a simpler, more conservative approach is often used, which sets the degrees of freedom to the smaller of

n_1 - 1

and

n_2 - 1

. Here,

n_1 = 15

and

n_2 = 12

, so the degrees of freedom are

\min(15 - 1, 12 - 1) = \min(14, 11) = 11

Question 10

A researcher performs a multiple linear regression to predict a student's final exam score based on three predictor variables: hours studied, number of missed classes, and prior GPA. The analysis is based on data from a class of 30 students. What are the degrees of freedom for the t-test used to determine if the coefficient for 'hours studied' is significantly different from zero?

26 (correct answer)
28
29
2

Explanation: In multiple linear regression, the degrees of freedom for the t-test of an individual coefficient are the residual or error degrees of freedom. This is calculated as

df = n - k - 1

, where

n

is the number of observations and

k

is the number of predictor variables. In this scenario,

n=30

and

k=3

. Therefore, the degrees of freedom are

30 - 3 - 1 = 26

Question 11

In which of the following scenarios is the use of a one-sample t-procedure for inference about a population mean most likely to produce misleading results?

A sample of size $n=50$ is taken from a population distribution that is moderately skewed to the right.
A sample of size $n=12$ is taken from a population that is known to be bimodal.
A sample of size $n=10$ is taken from a known normal population, but the sample contains one extreme outlier.
A sample of size $n=8$ is taken from a population that is strongly skewed, and the sample histogram also shows strong skew. (correct answer)

Explanation: The t-procedures are based on the assumption that the data are a random sample from a normal population. These procedures are robust to violations of normality, especially when the sample size is large. However, they perform poorly for very small sample sizes from populations that are strongly non-normal. Scenario D represents the worst combination: a very small sample size (

n=8

) from a strongly skewed population. In this case, the sampling distribution of the sample mean will not be close to normal, making the t-procedure inappropriate and likely to yield misleading p-values or confidence intervals.

Question 12

A researcher first calculates a 90% confidence interval for a mean using a sample of size $n=12$ . Subsequently, the researcher decides to increase the rigor of the study and calculates a 95% confidence interval using a new, larger sample of size $n=25$ . How will the critical value $t^*$ used for the second interval compare to the critical value used for the first?

The relationship cannot be determined without knowing the sample standard deviations.
The second $t^*$ will be smaller than the first.
The second $t^*$ will be identical to the first.
The second $t^*$ will be larger than the first. (correct answer)

Explanation: When comparing confidence intervals, you need to understand how critical values from the t-distribution respond to two key factors: confidence level and sample size (which determines degrees of freedom). The correct answer is D because two opposing forces are at work here. First, increasing the confidence level from 90% to 95% requires a larger critical value to capture more of the distribution's area. Second, increasing the sample size from n=12 to n=25 increases the degrees of freedom (from 11 to 24), which typically decreases the critical value as the t-distribution approaches the standard normal distribution. To determine which effect dominates, consider the actual values:

t_{0.05, 11} \approx 1.796

for the first interval and

t_{0.025, 24} \approx 2.064

for the second. The confidence level change (from 90% to 95%) has a stronger impact than the sample size increase, resulting in a larger critical value overall. Choice A is incorrect because critical values depend only on confidence level and degrees of freedom, not sample standard deviations. Choice B incorrectly assumes the sample size effect dominates over the confidence level effect. Choice C is wrong because these two different scenarios (different confidence levels and sample sizes) cannot yield identical critical values. Study tip: Remember that higher confidence always requires larger critical values, while larger sample sizes generally require smaller critical values. When both change simultaneously, calculate or look up the actual values to see which effect wins—don't assume one always dominates the other.

Question 13

A materials scientist tests the tensile strength of a new alloy. From a sample of 16 specimens, a 95% confidence interval for the mean tensile strength is calculated to be [345.2, 358.8] MPa. The scientist now wishes to test the hypothesis $H_0: \mu = 360$ MPa against the alternative $H_A: \mu \neq 360$ MPa. What can be concluded about the p-value for this test?

The p-value is less than 0.05. (correct answer)
The p-value is greater than 0.05.
The p-value is exactly 0.05.
The p-value cannot be determined without the sample mean and standard deviation.

Explanation: There is a direct duality between a

100(1-\alpha)\%

confidence interval and a two-sided hypothesis test at the

\alpha

significance level. If the value specified in the null hypothesis falls outside the confidence interval, then the null hypothesis would be rejected at that significance level. In this case, the null value

\mu = 360

is not contained within the 95% confidence interval [345.2, 358.8]. This is equivalent to rejecting

H_0

at the

\alpha = 0.05

level. Rejecting

H_0

means that the p-value is less than the significance level, so

p < 0.05

Question 14

A researcher constructs a 95% confidence interval for a population mean based on a sample of size $n=10$ . If a new sample of size $n=40$ is collected from the same population, what is the most accurate description of the change in the width of the 95% confidence interval, assuming the sample standard deviation $s$ is approximately the same for both samples?

The width will be approximately halved.
The width will be reduced by slightly more than half. (correct answer)
The width will be approximately quartered.
The width will be reduced by slightly less than half.

Explanation: The width of a t-confidence interval is given by

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

. When the sample size

n

is quadrupled (from 10 to 40), the

\sqrt{n}

term in the denominator doubles, which by itself would halve the width. However, the critical value

t^*

also changes because the degrees of freedom increase from

df_1 = 10-1 = 9

df_2 = 40-1 = 39

. As df increases, the t-distribution becomes more like the normal distribution, and the critical value

t^*

for a fixed confidence level decreases. For 95% confidence,

t^*_{df=9} \approx 2.262

and

t^*_{df=39} \approx 2.023

. Since both the standard error term (

s/\sqrt{n}

) and the critical value (

t^*

) decrease, the total width will be reduced by slightly more than half.

Question 15

Using data from a single random sample of size 22, an analyst computes two confidence intervals for the population mean. Interval 1 is [45.0, 55.0] and Interval 2 is [43.2, 56.8]. Which pair represents the confidence levels used for Interval 1 and Interval 2, respectively?

90% and 99%
99% and 95%
95% and 99% (correct answer)
95% and 90%

Explanation: When you encounter confidence interval problems, remember that confidence level and interval width are directly related: higher confidence levels produce wider intervals because you need to cast a broader net to be more certain you've captured the true population mean. Looking at the two intervals from the same sample, Interval 1 [45.0, 55.0] has a width of 10 units, while Interval 2 [43.2, 56.8] has a width of 13.6 units. Since Interval 2 is wider, it must correspond to the higher confidence level. Among the answer choices, only option C correctly identifies Interval 1 as 95% confidence and Interval 2 as 99% confidence. Let's examine why the other options fail: Option A suggests 90% and 99%, but this would mean the narrower interval (Interval 1) uses an even lower confidence level than 95%, which doesn't match the typical confidence levels used in practice. Option B proposes 99% and 95%, which incorrectly assigns the higher confidence level to the narrower interval—this violates the fundamental relationship between confidence and width. Option D suggests 95% and 90%, meaning the wider interval has lower confidence, which is mathematically impossible. The key insight is that when you're more confident about capturing the true mean (99% vs. 95%), you must accept a wider interval as the trade-off. Think of it like casting a fishing net: to be more certain of catching fish, you need a bigger net. Study tip: Always remember that confidence level and interval width move in the same direction—higher confidence always means wider intervals when using the same data.

Question 16

Consider a family of t-distributions indexed by their degrees of freedom, $v$ . Which of the following quantities is constant for all t-distributions, regardless of the value of $v$ (for $v > 1$ )?

The kurtosis of the distribution.
The value of the probability density function at 0, $f(0)$ .
The median of the distribution. (correct answer)
The variance of the distribution.

Explanation: All t-distributions are symmetric about 0. For any symmetric distribution, the mean, median, and mode are all equal. The mean of the t-distribution is 0 (for

v>1

), and thus the median is also 0 for all

v>1

. The other quantities change with

v

: The variance is

v/(v-2)

, which depends on

v

. The kurtosis is

6/(v-4) + 3

(for

v>4

), which depends on

v

. The height of the probability density function at 0,

f(0)

, increases as

v

increases. Therefore, only the median is constant.

Question 17

A researcher conducts a two-sample t-test assuming equal variances based on two independent samples of size $n_1=8$ and $n_2=10$ . Later, it is discovered that the two samples were not independent but were actually paired observations from 8 subjects measured twice. How does the correct number of degrees of freedom compare to the degrees of freedom used in the initial, incorrect analysis?

The correct degrees of freedom are 8, which is less than the 16 used initially.
The correct degrees of freedom are 7, which is less than the 18 used initially.
The correct degrees of freedom are 7, which is less than the 16 used initially. (correct answer)
The correct degrees of freedom are 16, which is the same as what was used initially.

Explanation: When you encounter a question about degrees of freedom in t-tests, the key is recognizing that the test type fundamentally changes the degrees of freedom calculation, not just the sample sizes involved. Let's work through what happened here. Initially, the researcher incorrectly performed a two-sample t-test with independent samples. For a two-sample t-test assuming equal variances, the degrees of freedom equal

(n_1 + n_2 - 2) = 8 + 10 - 2 = 16

. However, when the data is actually paired (the same 8 subjects measured twice), you must use a paired t-test instead. For a paired t-test, you're analyzing the differences between paired observations, effectively creating a single sample of differences. The degrees of freedom for a paired t-test equals the number of pairs minus 1. Since there are 8 subjects (8 pairs), the correct degrees of freedom is

8 - 1 = 7

. Looking at the wrong answers: Choice A correctly identifies 7 as less than 16, but incorrectly states the original degrees of freedom as 8. Choice B correctly calculates both degrees of freedom (7 and 16) but incorrectly claims the original analysis used 18 degrees of freedom. Choice D incorrectly suggests the degrees of freedom remain the same at 16. Remember this pattern: paired data always reduces your degrees of freedom compared to treating the same data as independent samples. When you see "paired" or "repeated measures," immediately think about whether the analysis method matches the data structure, as this directly impacts the degrees of freedom calculation.

Question 18

1.50 (correct answer)
1.25
1.20
1.00

Explanation: The t-distribution used for inference on a single mean has degrees of freedom

df = n-1

. In this case,

df = 7-1 = 6

. The theoretical variance of a t-distribution with

v

degrees of freedom is given by the formula

\text{Var}(T) = \frac{v}{v-2}

, which is defined for

v > 2

. Plugging in

v=6

, the variance is

\frac{6}{6-2} = \frac{6}{4} = 1.50

Question 19

11 (correct answer)
14
25
26

n_1 - 1

and

n_2 - 1

. Here,

n_1 = 15

and

n_2 = 12

, so the degrees of freedom are

\min(15 - 1, 12 - 1) = \min(14, 11) = 11

Question 20

24 (correct answer)
25
48
49

n-1

, where

n

is the number of pairs. Here, there are 25 students (pairs), so the degrees of freedom are

25 - 1 = 24