business statistics

Question

Test the claim that the mean GPA of night students is significantly different than the mean GPA of day students at the 0.2 significance level.

The null and alternative hypothesis would be:

H0:μN≤μDH0:μN≤μD

H1:μN>μDH1:μN>μD

H0:μN≥μDH0:μN≥μD

H1:μN<μDH1:μN<μD

H0:pN≤pDH0:pN≤pD

H1:pN>pDH1:pN>pD

H0:pN≥pDH0:pN≥pD

H1:pN<pDH1:pN<pD

H0:μN=μDH0:μN=μD

H1:μN≠μDH1:μN≠μD

H0:pN=pDH0:pN=pD

H1:pN≠pDH1:pN≠pD

The test is:right-tailed

two-tailed

left-tailed

The sample consisted of 40 night students, with a sample mean GPA of 3.03 and a standard deviation of 0.02, and 40 day students, with a sample mean GPA of 3.02 and a standard deviation of 0.06.

The test statistic is:  _____ (to 2 decimals)

The p-value is:  ______ (to 2 decimals)

Based on this we:

1. Fail to reject the null hypothesis
2. Reject the null hypothesis

Answer 1

Hi Salma,

To the first question, you have a mean GPA, not a proportion, so that eliminates choices c,e, and f. Null hypothesis means no difference, which implies an equals sign. We also only want to know about difference, not greater than/less than, so only option that works is: H0: muN=muD, Ha: muN not equal to muD. Now, we don’t know population standard deviation, so we have to use 2-sample t. Standard deviation for night is three times that for day, so we cannot pool. Therefore, formula for t-test statistic is:

t=[(xbar1-xbar2)-0]/sqrt[(s1²/n1) + (s2²/n2)]

where:

xbar1=sample mean 1=3.03

xbar2=sample mean 2=3.02

s1=standard deviation sample 1 (night) =0.02

s2=standard deviation sample 2 (day) =0.06

n1=sample size 1=40

n2=sample size 2=40

t=(3.03-3.02)/sqrt[(0.02²/40) + (0.06²/40)]

t=1.00

I will use the conservative route this time—we can use the smaller of n1-1 or n2-1 to compute degrees of freedom. Alternatively, we can use the Satterthwaite approximation, but that computation is quite intense. For our purposes, we’ll assume:

df=n1 or n2-1=40-1=39

Closest value to 39 on t-table without going over is 30. Looking across the row for 30 degrees of freedom, 1.00 falls between 0.854 and 1.055. Corresponding 2-sided p-values give:

0.30 < p < 0.40

That’s the closest estimate of p-value we can get without statistical software or the cumbersome Satterthwaite. If p is not correct, textbook author or instructor likely used such software or the Satterthwaite. With a p-value like, this, we fail to reject the null and conclude that true mean GPA of night students does not differ significantly from true mean GPA of day students.