business statistics

Question

Test the claim that the mean GPA of night students is larger than the mean GPA of day students at the 0.025 significance level.

The null and alternative hypothesis would be:

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:μN=μDH0:μN=μD

H1:μN≠μDH1:μN≠μD

H0:pN≥pDH0:pN≥pD

H1:pN<pDH1:pN<pD

H0:μN≥μDH0:μN≥μD

H1:μN<μDH1:μN<μD

The test is:left-tailed

right-tailed

two-tailed

The sample consisted of 30 night students, with a sample mean GPA of 2.76 and a standard deviation of 0.08, and 30 day students, with a sample mean GPA of 2.73 and a standard deviation of 0.04.

The test statistic is:  (to 2 decimals)

The p-value is:  (to 2 decimals)

Based on this we:

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

Answer 1

Hi Salma,

To the first part of your question, you are working with means here, not proportions, so you can eliminate any answers that do not include means. Your question also stipulates night students having greater mean GPAs than day, so you can also eliminate any answers that do not include "greater than" in the alternative hypothesis. Only answer that meets both criteria is:

H0:μN≤μDH0:μN≤μD

H1:μN>μDH1:μN>μD

Additionally, greater than implies a right-tailed test.

Now, we know nothing about the population standard deviation, which means we have to use t, not z. Our sample standard deviations are 0.02 and 0.04, so we just barely meet criteria for pooled t. If one sample standard deviation is more than twice the other, we would have to use a different t-test, which involves more intense calculations. As it is, we can use pooled t:

t=(x-bar1-x-bar2)/sqrt[s_p²((1/n₁) + (1/n₂))]

x-bar₁=night sample mean

x-bar₂=day sample mean

s_p=pooled standard deviation

n₁=night sample size

n₂=day sample size

We also need to compute the pooled standard deviation, which is:

s_p=sqrt[((n₁-1)(s₁²)) + ((n₂-1)(s₂²))/(n₁+n₂-2)]

n₁₌night sample size

n₂=day sample size

s₁=night sample standard deviation

s₂=day sample standard deviation

For this problem:

n₁=30

n₂=30

s₁=0.08

s₂=0.04

s_p=sqrt[((n₁-1)(s₁²)) + ((n₂-1)(s₂²))/(n₁+n₂-2)]

s_p=sqrt[(29*0.08²) + (29*0.04²)]/(30 + 30 -2)]

s_p=0.063

Now, back to our t-test:

t=(x-bar1-x-bar2)/sqrt[s_p²((1/n₁) + (1/n₂))]

x-bar₁=2.76

x-bar₂=2.73

s_p=0.063

n₁=30

n₂=30

t=(2.76-2.73)/sqrt(0.063²((1/30) + (1/30))

t=1.84 This is your test statistic.

Now, to get a p-value approximation, we need degrees of freedom. Formula for that with pooled t is:

df= n₁ + n₂ - 2

Thus:

df= 30 + 30 - 2

df=58

On our t-table, we have to do this Price is Right style--closest without going over. On my table, that is 40 degrees of freedom. Now, looking across that row, our test statistic of 1.84 falls between 1.684 and 2.021. Now, we have a one-sided test here--remember it was the greater than alternative--so check corresponding p-values near the top:

0.025 < p < 0.05

This is the most accurate we can get without statistical software. Now, your significance level was 0.025, and it is possible for p to exceed this based on the above, so we fail to reject H₀ and conclude no significant difference in mean GPA between night and day students.

I hope this helps.