Questions: students are shown below. selected female statistics Female: 95 Assume both follow a Normal distribution. What can be concluded the the α=0.10 level of significance level of significance? For this study, we should use t-test for the difference between two independent population means a. The null and alternative hypotheses would be: H0 : μ1 > μ2 H1 : μ1 ≠ μ2 b. The test statistic = -0.989 (please show your answer to 3 decimal places.) c. The p-value = 0.3367 (Please show your answer to 4 decimal places.)

students are shown below.
selected female statistics

Female: 95
Assume both follow a Normal distribution. What can be concluded the the α=0.10 level of significance level of significance?

For this study, we should use t-test for the difference between two independent population means
a. The null and alternative hypotheses would be:
H0 : μ1 > μ2
H1 : μ1 ≠ μ2
b. The test statistic = -0.989 (please show your answer to 3 decimal places.)
c. The p-value = 0.3367 (Please show your answer to 4 decimal places.)

Transcript text: students are shown below. selected female statistics Female: 95 Assume both follow a Normal distribution. What can be concluded the the $\alpha=0.10$ level of significance level of significance? For this study, we should use $t$-test for the difference between two independent population means a. The null and alternative hypotheses would be: $H_{0}$ : $\mu 1$ $\square$ $>$ $\mu 2$ $\square$ $\square$ 0 $H_{1}$ : $\mu 1$ ‡ $\mu 2$ $\square$ (Please enter a decimal) $\square$ $\infty$ 0 b. The test statistic $\square$ $\square$ $-0.989$ (please show your answer to 3 decimal places.) $\square$ c. The $p$-value $=$ $\square$ 0.3367 (Please show your answer to 4 decimal places.)

Solution

Solution Steps

Step 1: Hypotheses

We set up the null and alternative hypotheses as follows: \[ H_0: \mu_1 = \mu_2 \] \[ H_1: \mu_1 \neq \mu_2 \]

Step 2: Standard Error Calculation

The standard error $ SE $ is calculated using the formula: \[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{10.0}{5} + \frac{10.0}{5}} = 2.0 \]

Step 3: Test Statistic Calculation

The test statistic $ t $ is computed as: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{96.0 - 92.0}{2.0} = 2.0 \]

Step 4: Degrees of Freedom Calculation

The degrees of freedom $ df $ are calculated using the formula: \[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} = \frac{16.0}{2.0} = 8.0 \]

Step 5: P-value Calculation

The p-value $ P $ is calculated as: \[ P = 2(1 - T(|t|)) = 2(1 - T(2.0)) = 0.0805 \]

Final Answer

a. The null and alternative hypotheses are: \[ H_0: \mu_1 = \mu_2, \quad H_1: \mu_1 \neq \mu_2 \]
b. The test statistic is: \[ t = 2.0 \]
c. The p-value is: \[ P = 0.0805 \]

Thus, the final boxed answers are: \[ \boxed{H_0: \mu_1 = \mu_2, \quad H_1: \mu_1 \neq \mu_2} \] \[ \boxed{t = 2.0} \] \[ \boxed{P = 0.0805} \]

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