The mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an averageof three English courses with a standard deviation of 0.8. The females took an average off our English courses with a standard deviation of 1.0.

(a) Are the means statistically the same?

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let m be the subscript for the number of English courses taken by male students and f be the subscript for the number of English courses taken by female students.

Therefore, the population means would be μm and μf.

The random variable is xm - xf = difference in the sample mean number of English courses taken by male and female students.

We would set up the hypothesis.

The null hypothesis is

H0 : μm = μf H0 : μm - μf = 0

The alternative hypothesis is

H1 : μm ≠ μf H1 : μm - μf ≠ 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(xm - xf) / √ (sm²/nm + sf²/nf)

From the information given,

xm = 3

xf = 4

sm = 0.8

sf = 1

nm = 29

nf = 16

t = (3 - 4) / √ (0.8²/29 + 1²/16)

t = - 11.82

The formula for determining the degree of freedom is

df = [sm²/nm + sf²/nf]² / (1/nm - 1) (sm²/nm) ² + (1/nf - 1) (sf²/nf) ²

df = [0.8²/29 + 1²/16]²/[ (1/29 - 1) (0.8²/29) ² + (1/16 - 1) (1²/16) ²] = 0.00715/0.00027781093

df = 25.7

Approximately, df = 26

We would determine the probability value from the t test calculator. It becomes

p value = 0.00001

Assuming a level of significance of 0.05, we would reject the null hypothesis because the p value, 0.00001 is < 0.05

Therefore, we can conclude that the means are not statistically the same.