A student researcher compares the heights of men and women from the student body
ID: 3223276 • Letter: A
Question
A student researcher compares the heights of men and women from the student body of a certain college in order to estimate the difference in their mean heights. A random sample of 12 men had a mean height of 70.7inches with a standard deviation of 2.41 inches. A random sample of 17women had a mean height of 62.7inches with a standard deviation of3.07 inches. Determine the 98% confidence interval for the true mean difference between the mean height of the men and the mean height of the women. Assume that the population variances are equal and that the two populations are normally distributed.
Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.
Find the standard error of the sampling distribution to be used in constructing the confidence interval. Round your answer to two decimal places.
Construct the 98%confidence interval. Round your answers to two decimal places.
Explanation / Answer
Critical value:
Degrees of freedom = n1 + n2 - 2 = 12 + 17 - 2 = 27
For 27 degrees of freedom and 98% confidence interval,
Critical t value = 2.473
The statistical output for 98% confidence interval is:
Two sample T confidence interval:
1 : Mean of Population 1
2 : Mean of Population 2
1 - 2 : Difference between two means
(with pooled variances)
98% confidence interval results:
Hence,
Standard error = 1.06
98% confidence interval: (5.37, 10.63)
Difference Sample Diff. Std. Err. DF L. Limit U. Limit 1 - 2 8 1.0631766 27 5.3711258 10.628874Related Questions
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