CI for a Mean - College Statistics

A researcher calculates a 95% confidence interval for the mean height of a plant species based on a sample of 20 plants. The interval is (30 cm, 38 cm). Later, it is discovered that one plant's height was recorded as 5 cm, but it should have been 25 cm. All other data points were between 28 cm and 40 cm. How will correcting this single data entry error affect the confidence interval?
Answer: The new interval will be narrower and its center will be higher. The original value of 5 cm was a significant low outlier. Correcting it to 25 cm will have two main effects. First, it will increase the sample mean ( $\bar{x}$ ), shifting the center of the confidence interval to the right (higher). Second, the original outlier would have greatly inflated the sample standard deviation (s). Replacing the outlier with a value closer to the mean will decrease the sample standard deviation. A smaller s results in a smaller margin of error and thus a narrower confidence interval. Therefore, the new interval will be both narrower and centered at a higher value.
A 95% confidence interval for a population mean, based on a random sample of size 16, is calculated to be (22.1, 27.9). Assuming all conditions for the t-interval are met, what is the sample standard deviation?
Answer: 5.44 The sample mean ( $\bar{x}$ ) is the midpoint of the interval: $\bar{x} = (22.1 + 27.9) / 2 = 25.0$ . The margin of error (ME) is half the width of the interval: $ME = 27.9 - 25.0 = 2.9$ . The formula for the margin of error is $ME = t^* \frac{s}{\sqrt{n}}$ . For a 95% confidence interval with $n=16$ , the degrees of freedom are $df = 16 - 1 = 15$ . The critical value $t^*$ for df=15 is 2.131. We can solve for the sample standard deviation (s): $2.9 = 2.131 \times \frac{s}{\sqrt{16}}$ , which simplifies to $2.9 = 2.131 \times \frac{s}{4}$ . Solving for s gives $s = \frac{2.9 \times 4}{2.131} \approx 5.44$ .
A biologist measures the length of 15 randomly sampled trout from a lake. The data are summarized in the boxplot. The biologist wishes to construct a 95% confidence interval for the mean length of all trout in the lake. Based on the provided boxplot, what is the most significant concern about the validity of using a t-interval?
Answer: The sample data show strong skewness and contain an outlier with a small sample size, making the normality condition for the t-procedure questionable. The t-interval procedure requires that the data come from a nearly normal population or that the sample size is large enough (typically n > 30) for the Central Limit Theorem to apply. With a small sample size (n=15), the procedure is not robust to strong skewness or outliers. The boxplot indicates both left-skewness and a low outlier, violating the 'nearly normal' condition. Choice A is incorrect; a z-interval requires the population standard deviation to be known. Choice C mentions a valid alternative (interval for the median) but doesn't identify the primary reason the t-interval itself is invalid. Choice D contradicts the stem, which states the sample was random.
A quality control specialist constructs a 99% confidence interval for the mean weight of cereal boxes, which is found to be (15.8 oz, 16.3 oz). The company's stated net weight is 16 oz. Based only on this interval, which of the following is a valid conclusion regarding a two-sided hypothesis test of $H_0: \mu = 16$ versus $H_a: \mu \neq 16$ ?
Answer: The null hypothesis would not be rejected at the $\alpha = 0.01$ significance level. A confidence interval contains a range of plausible values for the population parameter. There is a direct correspondence between a two-sided hypothesis test and a confidence interval: if the null value ( $\mu_0 = 16$ ) is contained within the C% confidence interval, then the null hypothesis would not be rejected at the $\alpha = 1 - C/100$ significance level. Here, 16 is within the 99% confidence interval (15.8, 16.3). Therefore, we would not reject $H_0: \mu = 16$ at the $\alpha = 1 - 0.99 = 0.01$ level. This also means the p-value must be greater than 0.01. The sample mean is the center of the interval, (15.8 + 16.3)/2 = 16.05 oz, not 16 oz.
A researcher is creating a confidence interval for the mean score on a standardized test where scores can only range from 0 to 100. Which of the following calculated 95% t-intervals is impossible to obtain from any real sample of data, regardless of the sample size or standard deviation?
Answer: (102, 112) The center of a confidence interval is the sample mean, $\bar{x}$ . For a test with scores from 0 to 100, any individual score must be in this range, and therefore the mean of any sample of scores, $\bar{x}$ , must also be within the range [0, 100]. Let's examine the centers of the given intervals: A) Center = 0, possible. B) Center = 100, possible. C) Center = (102+112)/2 = 107, impossible. D) Center = 50, possible. Since it is impossible to have a sample mean of 107 for data that cannot exceed 100, the interval (102, 112) could not have been generated from a valid sample. While intervals A and B have endpoints outside the [0, 100] range, this is possible if the sample mean is close to a boundary and the margin of error is sufficiently large.
To test a new fuel additive, a researcher measures the gas mileage (in mpg) of 12 cars both before and after the additive is used. The researcher plans to construct a 95% confidence interval to estimate the effectiveness of the additive. Which of the following is the correct parameter for which the confidence interval should be constructed?
Answer: The population mean of the paired differences in gas mileage (mileage after - mileage before) for all cars of this type. This is a paired data design because two measurements (before and after) are taken on each experimental unit (each car). The proper way to analyze paired data is to first calculate the difference for each pair (e.g., mileage_after - mileage_before). Then, a one-sample t-interval is constructed for the mean of these differences. This parameter, the population mean of the paired differences ( $\mu_{diff}$ ), represents the average effect of the additive. A two-sample procedure (A) is incorrect as it assumes the two groups of measurements are independent, which they are not. Analyzing only the 'after' data (B) ignores the baseline comparison. The standard deviation (D) is used in the calculation but is not the parameter being estimated.
To estimate the average time students at a large university spend on a new learning module, a professor collects data from all 35 students in her advanced statistics class. She then calculates a 95% t-interval for the mean time. What is the most significant flaw in this statistical procedure?
Answer: The sample is not representative of the intended population, making inference to all university students inappropriate. The primary issue is that the sample is a convenience sample, not a random sample from the population of interest (all students at the university). Students in an advanced statistics class are likely not representative of the general student body in terms of study habits, motivation, or background knowledge. Therefore, any inference made from this sample to the entire university population is likely to be biased and invalid. While the CLT (B) might apply to the sampling distribution shape, it does not correct for a biased sampling method. Choice (C) is true for the class, but the goal was to infer to the university. Choice (D) is incorrect; a t-interval is precisely the procedure for when the population standard deviation is unknown.
A researcher wants to construct an 80% confidence interval for a population mean based on a random sample of size $n=18$ . Which of the following is the correct critical value $t^*$ to use?
Answer: 1.333 For an 80% confidence interval, the total area in the two tails of the t-distribution is $\alpha = 1 - 0.80 = 0.20$ . For a two-sided interval, we need the area in each tail to be $\alpha/2 = 0.10$ . The degrees of freedom are $df = n - 1 = 18 - 1 = 17$ . We need to find the t-value with 17 degrees of freedom that has 0.10 area to its right. Using a t-table or calculator, this value is $t_{0.10, 17} = 1.333$ . Choice B uses df=18. Choice C uses a tail area of 0.20 instead of 0.10. Choice D is the z* critical value for 80% confidence.
A 90% confidence interval for a mean is calculated from a sample of size $n=25$ . If a new sample of size $n=100$ is taken from the same population, and by coincidence the new sample mean and sample standard deviation are identical to those of the first sample, how would the width of the new 90% confidence interval compare to the original interval's width?
Answer: The new interval's width would be slightly less than half the original width. The width of a t-interval is $2 \times t^* \times \frac{s}{\sqrt{n}}$ . Quadrupling the sample size from 25 to 100 makes the $\frac{1}{\sqrt{n}}$ term half as large ( $\frac{1}{\sqrt{100}} = \frac{1}{10}$ is half of $\frac{1}{\sqrt{25}} = \frac{1}{5}$ ). However, the critical value $t^*$ also changes because the degrees of freedom increase. For a 90% CI, $t^*$ with $df=24$ is 1.711. For $df=99$ , $t^*$ is 1.660. Since the new $t^*$ is smaller, the new width will be $\frac{1.660}{1.711} \approx 0.97$ times half the original width. This results in a new interval that is slightly less than half the width of the original.
A study with a sample size of $n=25$ produced a confidence interval with a margin of error of 4.0. The sample standard deviation was $s=10.0$ . What is the approximate confidence level of this interval?
Answer: 95% First, calculate the standard error of the mean: $SE = s / \sqrt{n} = 10.0 / \sqrt{25} = 10.0 / 5 = 2.0$ . The margin of error formula is $ME = t^* \times SE$ . We can solve for the critical value $t^*$ : $4.0 = t^* \times 2.0 \Rightarrow t^* = 2.0$ . We need to find the confidence level corresponding to a critical value of $t^*=2.0$ with degrees of freedom $df = n - 1 = 24$ . Looking at a t-distribution table for df=24, a $t^*$ of 1.711 corresponds to a 90% confidence level, and a $t^*$ of 2.064 corresponds to a 95% confidence level. Since our calculated $t^*$ of 2.0 is very close to 2.064, the confidence level is approximately 95%.
A biologist measures the weights of a random sample of 25 adult squirrels and computes a 90% confidence interval for the mean weight to be (480.5, 519.5) grams. If the weights had been recorded in kilograms instead of grams (1 kg = 1000 g), what would the 90% confidence interval have been?
Answer: (0.4805, 0.5195) When changing units of measurement, all summary statistics based on those units will change by the same conversion factor. To convert from grams to kilograms, we divide by 1000. This applies to the sample mean, the standard deviation, and consequently, the endpoints of the confidence interval. The lower bound becomes $480.5 / 1000 = 0.4805$ kg, and the upper bound becomes $519.5 / 1000 = 0.5195$ kg. The resulting interval is (0.4805, 0.5195).
A 95% confidence interval for a population mean is calculated from a sample of size $n=10$ . The sample standard deviation is $s=5$ . If the population standard deviation $\sigma$ was actually known to be exactly 5, how would a 95% confidence interval calculated using $\sigma$ (a z-interval) compare to the original t-interval?
Answer: The interval calculated using $\sigma$ would be narrower. The width of the interval is determined by the critical value. The original t-interval uses a critical value $t^*$ from the t-distribution with $n-1=9$ degrees of freedom. For 95% confidence, $t^*_{df=9} = 2.262$ . An interval calculated with a known $\sigma$ would use a critical value $z^*$ from the standard normal distribution. For 95% confidence, $z^* = 1.96$ . Since $z^*=1.96$ is less than $t^*=2.262$ , and all other components (s= $\sigma$ , n) are the same, the margin of error for the z-interval would be smaller, resulting in a narrower interval. The t-distribution has heavier tails than the normal distribution to account for the extra uncertainty of estimating $\sigma$ with s, which always makes $t^* > z^*$ for any given confidence level.
An environmental scientist wants to estimate the mean level of a certain pollutant in a river to within 0.5 parts-per-million (ppm) with 95% confidence. A pilot study of 12 samples yielded a standard deviation of 2.3 ppm. What is the minimum number of additional samples the scientist must take to achieve the desired margin of error?
Answer: 70 To calculate the required sample size $n$ for a given margin of error (ME), we use the formula $n = (\frac{z^* \cdot s}{ME})^2$ . We use the z* critical value (1.96 for 95% confidence) as a conservative estimate because the t* value depends on the unknown $n$ . We use the standard deviation from the pilot study (s=2.3) as an estimate. Plugging in the values: $n = (\frac{1.96 \times 2.3}{0.5})^2 = (9.016)^2 \approx 81.29$ . Since the sample size must be an integer, we must round up to $n=82$ to ensure the margin of error is not exceeded. The question asks for the number of additional samples. Since 12 were already taken in the pilot study, the additional number needed is $82 - 12 = 70$ .
A researcher plans to take two independent random samples from the same large population. Sample A will have 25 observations and Sample B will have 100 observations. Assume the sample standard deviation (s) turns out to be roughly the same for both samples. Which statement best describes the relationship between the expected standard error of the mean (SE) for the two samples?
Answer: The standard error for Sample B is expected to be about half the standard error for Sample A. The formula for the standard error of the mean is $SE = s/\sqrt{n}$ . For Sample A, $SE_A = s/\sqrt{25} = s/5$ . For Sample B, $SE_B = s/\sqrt{100} = s/10$ . Comparing the two, $SE_B = s/10 = (1/2) \times (s/5) = (1/2) \times SE_A$ . Thus, the standard error for the larger sample (B) is expected to be half that of the smaller sample (A). Choice A is incorrect because it treats the relationship as $1/n$ instead of $1/\sqrt{n}$ .
From the same population, Researcher 1 takes a random sample of size 40 and computes a 95% CI for the mean, getting (11.2, 14.8). Researcher 2 takes an independent random sample of size 60 and computes a 95% CI for the same population mean, getting (13.5, 15.5). Which of the following is the most appropriate conclusion?
Answer: Both intervals are plausible estimates for the true population mean, and the differences between them are expected due to sampling variability. Random samples from the same population will produce different sample statistics (like mean and standard deviation) and thus different confidence intervals. This is the essence of sampling variability. Both intervals were constructed with a method that is successful 95% of the time in the long run. Neither is definitively 'correct' or 'incorrect'. The fact that they are different is expected. Choice A makes too strong a claim; while the overlap region is highly plausible, the true mean could be in one interval but not the other. Choice B misinterprets the 95% confidence level, which applies to the method for both researchers. Choice C incorrectly dismisses the expected role of random chance in sampling.
A 95% confidence interval for the mean number of hours college students study per week is found to be (12.5, 15.5). Which of the following is the most accurate interpretation of this interval?
Answer: If we were to repeatedly take random samples of the same size and construct an interval for each, 95% of those intervals would be expected to contain the true population mean. The confidence level (95%) refers to the long-run success rate of the method used to construct the interval, not the probability that a specific interval contains the true parameter. Choice A is a common misinterpretation; the true mean is a fixed value, so it is either in the interval or not (probability is 0 or 1). Choice B incorrectly interprets an interval for a population mean as an interval containing individual data points. Choice D incorrectly describes the interval as a range for future sample means rather than an estimate for the population mean.
The width of a 95% confidence interval for a population mean, based on a sample of size $n$ , is 10 units. If the study were repeated with a new sample of size $4n$ from the same population and the new sample standard deviation was identical to the first, what would be the approximate width of the new 95% confidence interval?
Answer: Slightly less than 5 units The width of a t-interval is $W = 2 \cdot t^* \cdot \frac{s}{\sqrt{n}}$ . The new width, $W_{new}$ , would be $2 \cdot t^*_{new} \cdot \frac{s}{\sqrt{4n}} = 2 \cdot t^*_{new} \cdot \frac{s}{2\sqrt{n}} = (\frac{t^*_{new}}{t^*_{old}}) \cdot (\frac{W_{old}}{2})$ . Increasing the sample size from $n$ to $4n$ increases the degrees of freedom. As degrees of freedom increase, the t-distribution gets closer to the normal distribution, and the critical value $t^*$ for a fixed confidence level decreases (gets closer to $z^*$ ). Therefore, the ratio $t^*_{new} / t^*_{old}$ will be slightly less than 1. This means the new width will be slightly less than half the old width, or slightly less than 5 units.
The distribution of household incomes in a large country is known to be strongly right-skewed. An economist takes a random sample of 40 households and finds that the distribution of the natural logarithm of the incomes, ln(income), is approximately symmetric and mound-shaped. A 95% confidence interval for the mean of the log-transformed incomes is calculated to be (10.5, 10.9). Which of the following is a valid interpretation based on this interval?
Answer: A 95% confidence interval for the median income in the population is approximately ( $e^{10.5}$ , $e^{10.9}$ ). When a transformation (like the natural log) makes the data symmetric, the mean of the transformed data is a good estimate of the median of the transformed data. Therefore, the interval (10.5, 10.9) is a 95% confidence interval for the population median of ln(income). Applying the inverse transformation (exponentiating) to the endpoints gives a confidence interval for the population median of the original variable, income. Choice A is incorrect because the mean of the logs is not the log of the mean ( $E[\ln(X)] \neq \ln[E(X)]$ ). Because the original distribution is right-skewed, the mean income will be greater than the median income.
A 99% confidence interval for a population mean is (55.4, 68.6). The sample standard deviation was $s=15.0$ . What was the sample size, $n$ ?
Answer: 38 First, find the sample mean and margin of error. $\bar{x} = (55.4 + 68.6)/2 = 62.0$ . $ME = 68.6 - 62.0 = 6.6$ . The formula is $ME = t^* \frac{s}{\sqrt{n}}$ , so $6.6 = t^* \frac{15}{\sqrt{n}}$ . Since $t^*$ depends on $n$ , we must iterate or test the answer choices. Let's test $n=38$ . If $n=38$ , then $df=37$ . The $t^*$ critical value for 99% confidence with df=37 is approximately 2.715. Let's see if this works: $ME = 2.715 \times \frac{15}{\sqrt{38}} \approx 2.715 \times \frac{15}{6.164} \approx 6.608$ . This is extremely close to the given ME of 6.6. Testing the other options: For n=35 (df=34), $t^*=2.728$ , ME = $2.728 \times 15/\sqrt{35} \approx 6.91$ . For n=21 (df=20), $t^*=2.845$ , ME = $2.845 \times 15/\sqrt{21} \approx 9.32$ . Thus, $n=38$ is the correct sample size.
A food scientist analyzed the vitamin C content (in mg) in a random sample of apples of a certain variety. The analysis yielded the computer output shown. Which of the following is the 95% confidence interval for the mean vitamin C content?
Answer: (9.64, 11.26) The formula for a confidence interval is $\bar{x} \pm t^* \times SE_{Mean}$ . From the output, $\bar{x} = 10.45$ , $SE_{Mean} = 0.388$ , and $n=22$ . The degrees of freedom are $df = n-1 = 21$ . For a 95% confidence level with df=21, the critical value is $t^* = 2.080$ . The margin of error is $ME = 2.080 \times 0.388 = 0.80704$ . The interval is $10.45 \pm 0.80704$ , which is (9.64296, 11.25704), or approximately (9.64, 11.26). Choice A uses z*=1.96. Choice B incorrectly uses the standard deviation (StDev) instead of the standard error (SE Mean). Choice D uses df=22 instead of df=21 for the critical value.

Start this quiz