Suppose there is a population of test scores on a large, standardized exam for which the mean and standard deviation are unknown. Two different random samples of 50 data values are taken from the population. One sample has a larger sample standard deviation than the other. Each of the samples is used to construct a 95% confidence interval. How do you think these two confidence intervals would compare?
Question
Suppose there is a population of test scores on a large, standardized exam for which the mean and standard deviation are unknown. Two different random samples of 50 data values are taken from the population. One sample has a larger sample standard deviation than the other. Each of the samples is used to construct a 95% confidence interval. How do you think these two confidence intervals would compare?
Solution
The two 95% confidence intervals would differ in their widths. The confidence interval constructed from the sample with the larger standard deviation would be wider than the one constructed from the sample with the smaller standard deviation.
Here's why:
-
The formula for a confidence interval is: sample mean ± (critical value * standard deviation / √n), where n is the sample size. The critical value is determined by the desired level of confidence (95% in this case).
-
The standard deviation is a measure of the dispersion or spread in a distribution. A larger standard deviation means that the data points are more spread out from the mean, and vice versa.
-
Therefore, if one sample has a larger standard deviation than the other, it means that the data points in that sample are more spread out. This increased spread is reflected in the confidence interval, which will be wider for the sample with the larger standard deviation.
-
Conversely, the sample with the smaller standard deviation has data points that are closer to the mean, resulting in a narrower confidence interval.
So, in conclusion, the confidence interval of the sample with the larger standard deviation would be wider, indicating a greater uncertainty about the population mean, compared to the sample with the smaller standard deviation.
Similar Questions
A sample of size =n92 is drawn from a normal population whose standard deviation is =σ5.1. The sample mean is =x35.33.Part 1 of 2(a) Construct a 99.9% confidence interval for μ. Round the answer to at least two decimal places.A 99.9% confidence interval for the mean is 33.58<<μ37.08.Part: 1 / 21 of 2 Parts CompletePart 2 of 2(b) If the population were not approximately normal, would the confidence interval constructed in part (a) be valid? Explain.The confidence interval constructed in part (a) ▼(Choose one) be valid since the sample size ▼(Choose one) larg
A study was conducted to estimate μ, the mean number of weekly hours that U.S. adults use computers at home. Suppose a random sample of 81 U.S. adults gives a mean weekly computer usage time of 8.5 hours and that from prior studies, the population standard deviation is assumed to be σ = 3.6 hours. The 95% confidence interval for the mean, μ, is (7.7, 9.3).Which of the following will provide a more informative (i.e., narrower) confidence interval than the 95% confidence interval? Check all that apply. Using a sample of size 400 (instead of 81) Using a sample of size 36 (instead of 81) Using a different sample of size 81 Using a 90% level of confidence (instead of 95%) Using a 99% level of confidence (instead of 95%)
Suppose you are to compare the average real estate value of houses in 2 cities. You collected samples from each city and compare their mean and variance. The data are given below. Group Sample size Sample mean ($000)Sample std. devCity A 37 230 17City B 40 200 15 Assuming unequal population variances and using the degrees of freedom of min(n1-1,n2-1), where n1 is the sample size of group 1 and n2 of group 2, calculate the 90% confidence interval of the difference between the means.
Which confidence level would produce the widest interval when estimating the mean of a population from the mean and standard deviation of a sample of that population?A.26%B.54%C.11%D.38%
A student was asked to find a 95% confidence interval for weight of their backpacks in pounds using data from a random sample of size n = 20. Which of the following is a correct interpretation of the interval 3.1 < μ < 8.6? Assume the population is normally distributed.With 95% confidence, the weight of a randomly selected backpack will be between 3.1 and 8.6 pounds.There is a 95% chance that the mean of a sample of 20 backpacks will weigh between 3.1 and 8.6 pounds.There is a 95% chance that the weight is between 3.1 and 8.6.With 95% confidence, the mean weight of all backpacks is between 3.1 and 8.6 pounds.The sample mean weight of all backpacks is between 3.1 and 8.6 pounds, 95% of the time. We know this is true because the mean of our sample is between 3.1 and 8.6.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.