Question

1.) Now use Minitab and attach "Output" for the problem below?

A random sample of 59 students was selected from the PSU student data base. Each student was asked their Sex and Height (in inches). Below are the summary results:

Sex	N	Mean	SD
Female	42	66.10	2.58
Male	17	70.12	3.66

a.) Using the summary information, estimate with 99% confidence the difference in heights between the sexes and interpret these results. What are the exact degrees of freedom used in the calculation of the t-statistic?

b.) Is there enough evidence at a 1% level of significance to conclude that males, on average, are taller than females? Include hypotheses, p-value, decision, and conclusion.

c.) What assumptions would need to be satisfied in order to safely use and interpret the results?

Answer 1

Solution:

Given data:

Female	Male
66.1	70.12
2.58	3.66
42	17

a) We have to find 99% confidence interval for the difference between height of female and height of male.

Following is the output:

	Female	Male
	66.1	70.12	mean
	2.58	3.66	std. dev.
	42	17	n
		57	df
		-4.0200	difference (Female - Male)

	-4.78 t		-6.2597	confidence interval 99.% lower
			-1.7803	confidence interval 99.% upper
			2.2397	margin of error

Hence, 99% confidence interval is,(-6.2597 , -1.7803)

b)

The null and alternative hypotheses:

The output is,

Decision: P-value is close to 0. Therefore, P-value <0.01 then reject H0.

Conclusion: There is enough evidence to conclude that males, on average, are taller than females at 0.01 level of significance.

c)

Here the given data is sample data which is in the form of summary. Therefore, we assume that this data is approximately normally distributed. In this scenario, population standard deviation is uknown and thus the t test for two mean is used. Also there is not given about equality of standard deviation, so we asumed it and followed pooled t-test.

Done