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Elan Coil [88]
3 years ago
10

The mean number of English courses taken in a two–year time period by male and female college students is believed to be about t

he 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?
Mathematics
1 answer:
prisoha [69]3 years ago
3 0

Answer:

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.

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