Question

test the claim that the mean gpa of night students is significantly different than the mean gpa of day students at the 0.1 significance level.
the null and alternative hypothesis would be:
$h_0:mu_n = mu_d$ $h_0:p_n = p_d$ $h_0:mu_n = mu_d$ $h_0:mu_n = mu_d$ $h_0:p_n = p_d$ $h_0:p_n = p_d$
$h_1:mu_nmu_d$ $h_1:mu_n
eqmu_d$ $h_1:p_n
eq p_d$ $h_1:p_n > p_d$
the test is:
two - tailed right - tailed left - tailed
the sample consisted of 75 night students, with a sample mean gpa of 2.85 and a standard deviation of 0.06, and 75 day students, with a sample mean gpa of 2.87 and a standard deviation of 0.08.
the test statistic is: (to 2 decimals)
the p - value is: (to 2 decimals)
based on this we:
reject the null hypothesis
fail to reject the null hypothesis
question help: message instructor post to forum

Explanation:

Step1: Identify null and alternative hypotheses

We are testing if the mean GPA of night students ($\mu_N$) is different from the mean GPA of day students ($\mu_D$). So, $H_0:\mu_N=\mu_D$ and $H_1:\mu_N
eq\mu_D$.

Step2: Determine the type of test

Since the alternative hypothesis has $
eq$, it is a two - tailed test.

Step3: Calculate the test statistic

The formula for the two - sample independent t - test statistic when population variances are unknown but assumed equal is $t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$, where $\bar{x}_1 = 2.85$, $\bar{x}_2=2.87$, $n_1 = n_2=75$, and the pooled standard deviation $s_p=\sqrt{\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}}$. First, calculate $s_p=\sqrt{\frac{(75 - 1)\times0.06^2+(75 - 1)\times0.08^2}{75+75 - 2}}\approx0.0707$. Then $t=\frac{(2.85 - 2.87)-0}{0.0707\sqrt{\frac{1}{75}+\frac{1}{75}}}\approx - 1.77$.

Step4: Calculate the p - value

The degrees of freedom is $df=n_1 + n_2-2=75 + 75-2 = 148$. Using a t - distribution table or calculator for a two - tailed test with $t=-1.77$ and $df = 148$, the p - value is approximately $0.078\approx0.08$.

Step5: Make a decision

Since the significance level $\alpha = 0.1$ and the p - value ($0.08$) is less than $\alpha$, we reject the null hypothesis.

Answer:

The null and alternative hypothesis: $H_0:\mu_N=\mu_D$, $H_1:\mu_N
eq\mu_D$
The test is: two - tailed
The test statistic: $-1.77$
The p - value: $0.08$
Based on this we: Reject the null hypothesis