Question

As part of a course project, a statistics student surveyed random samples of 50 student athletes and 50 student non-athletes at his university, with the goal of comparing the heights of the two groups. His summary statistics are displayed in the provided table.

	n		s
Athletes	50	68.96	4.25
Non-athletes	50	67.28	3.46

a). Which data analysis method is more appropriate in this situation: paired data difference in means or difference in means with two separate groups? Explain briefly.

b). Construct a 99% confidence interval for the difference in mean heights between student athletes and non-athletes at this university. Use two decimal places in your margin of error.

c). Test, at the 5% level, if student athletes at this university are significantly taller, on average, than student non-athletes. Include all of the details.

Answer 1

A) The un-paired test will be more preferrable for this situation. As both are students but divided into different independent groups athletes and non athletes

B) Following formula can be used to calculate the confidence intervals:

However, t value can be identified from the t-table at df (49,49) and at confidence interval of 99%

t value is 2.6799

Given Values:

n1 = 50

n2 = 50

s1 = 4.25

s2 = 3.46

X1 = 68.96

X2 = 67.28

By putting above values in the formula, we get:

Lower bound: (68.96 - 67.28) - 2.6799 * 0.775 = - 0.396

Upper Bound: (68.96 - 67.28) + 2.6799 * 0.775 = 3.756

C)

Null hypothesis: There is no difference heights of athlete and non-athlete students

Alternate Hypothesis: There is significant difference in heights of athletes and non-athlete students

Following formula can be used to calculate the test statistic:

Based on above formula, t value is 2.167

By referring to t table, we get p value = 0.031 at t=2.167 and df = 49.

Since the calculated p value (0.031) is more than the alpha value 0.01. the null hypothesis is accepted.

Therefore, it can be concluded that there is no difference between heights of athletes and non athletes.