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You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 30 night students, and the sample mean GPA is 2.35 with a standard deviation of 0.46. You sample 25 day students, and the sample mean GPA is 2.58 with a standard deviation of 0.47. Test the claim using a 5% level of significance. Assume the sample standard deviations are unequal and that GPAs are normally distributed. Give answer to exactly 4 decimal places.
Hypotheses:
sub(H,0):sub(μ,1) = sub(μ,2)
sub(H,1):sub(μ,1) â  sub(μ,2)
**I'm not sure how to calculate this in excel***
Enter the test statistic - round to 4 decimal places.
A=
Enter the p-value - round to 4 decimal places.
A=

Respuesta :

Answer:

Test statistic t = -1.8246

P-value = 0.0737

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean GPA of night students is different from the mean GPA of day students.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that the mean GPA of night students is different from the mean GPA of day students.

Then, the null and alternative hypothesis are:

[tex]H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0[/tex]

The significance level is 0.05.

The sample 1, of size n1=30 has a mean of 2.35 and a standard deviation of 0.46.

The sample 2, of size n2=25 has a mean of 2.58 and a standard deviation of 0.47.

The difference between sample means is Md=-0.23.

[tex]M_d=M_1-M_2=2.35-2.58=-0.23[/tex]

The estimated standard error of the difference between means is computed using the formula:

[tex]s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{0.46^2}{30}+\dfrac{0.47^2}{25}}\\\\\\s_{M_d}=\sqrt{0.007+0.009}=\sqrt{0.016}=0.1261[/tex]

Then, we can calculate the t-statistic as:

[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-0.23-0}{0.1261}=\dfrac{-0.23}{0.1261}=-1.8246[/tex]

The degrees of freedom for this test are:

[tex]df=n_1+n_2-2=30+25-2=53[/tex]

This test is a two-tailed test, with 53 degrees of freedom and t=-1.8246, so the P-value for this test is calculated as (using a t-table):

[tex]\text{P-value}=2\cdot P(t<-1.8246)=0.0737[/tex]

As the P-value (0.0737) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean GPA of night students is different from the mean GPA of day students.

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