Math / Data & Statistics

QuestionA personnel director in a particular state claims that the meanannual income is greater in one of the state's counties (county A) than it is in another county (county B). In County A, a random sample of 18 residents has a mean annual income of $\$ 40,800$ and a standard deviation of $\$ 8800$ . In County B, a random sample of 8 residents has a mean annual income of $\$ 37,600$ and a standard deviation of $\$ 5800$ . At $\alpha=0.10$ , answer parts (a) through (e). Assum the population variances are not equal. If convenient, use technology to solve the problem. D. "The mean annual incomes in counties $A$ and $B$ are not equal."
What are $\mathrm{H}_{0}$ and $\mathrm{H}_{\mathrm{a}}$ ? The null hypothesis, $H_{0}$ , is $\mu_{1} \leq \mu_{2}$ . The alternative hypothesis, $H_{a}$ , is $\mu_{1}>\mu_{2}$ . Which hypothesis is the claim? The null hypothesis, $\mathrm{H}_{0}$ The alternative hypothesis, $\mathrm{H}_{\mathrm{a}}$ (b) Find the critical value(s) and identify the rejection region(s).
Enter the critical value(s) below. 1.345 (Type an integer or decimal rounded to three decimal places as needed. Use a comma to separate answers as needed.)

Studdy Solution

STEP 1

1. The population variances are not equal.
2. The sample sizes are 18 for County A and 8 for County B.
3. The significance level $\alpha = 0.10$ .
4. The test is a one-tailed test since the claim is that the mean income in County A is greater than in County B.

STEP 2

1. Define the hypotheses.
2. Determine the appropriate test statistic and distribution.
3. Find the critical value(s) for the test.
4. Identify the rejection region(s).

STEP 3

Define the hypotheses:
- Null hypothesis ( $H_0$ ): $\mu_1 \leq \mu_2$ - Alternative hypothesis ( $H_a$ ): $\mu_1 > \mu_2$

STEP 4

Since the population variances are not equal and the sample sizes are small, use the Welch's t-test. The test statistic follows a t-distribution with degrees of freedom calculated using the Welch-Satterthwaite equation.

STEP 5

Find the critical value for a one-tailed test at $\alpha = 0.10$ .
- Use a t-distribution table or technology to find the critical value for the calculated degrees of freedom.
Given the problem statement, the critical value is:
$t_{critical} = 1.345$

STEP 6

Identify the rejection region:
- Since it is a one-tailed test, the rejection region is in the right tail of the t-distribution. - Reject $H_0$ if the test statistic is greater than $1.345$ .
The critical value is:
$\boxed{1.345}$

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Studdy Solution

STEP 1

1. The population variances are not equal.
2. The sample sizes are 18 for County A and 8 for County B.
3. The significance level $\alpha = 0.10$ .
4. The test is a one-tailed test since the claim is that the mean income in County A is greater than in County B.

STEP 2

1. Define the hypotheses.
2. Determine the appropriate test statistic and distribution.
3. Find the critical value(s) for the test.
4. Identify the rejection region(s).

STEP 3

Define the hypotheses:
- Null hypothesis ( $H_0$ ): $\mu_1 \leq \mu_2$ - Alternative hypothesis ( $H_a$ ): $\mu_1 > \mu_2$

STEP 4

Since the population variances are not equal and the sample sizes are small, use the Welch's t-test. The test statistic follows a t-distribution with degrees of freedom calculated using the Welch-Satterthwaite equation.

STEP 5

Find the critical value for a one-tailed test at $\alpha = 0.10$ .
- Use a t-distribution table or technology to find the critical value for the calculated degrees of freedom.
Given the problem statement, the critical value is:
$t_{critical} = 1.345$

STEP 6

Identify the rejection region:
- Since it is a one-tailed test, the rejection region is in the right tail of the t-distribution. - Reject $H_0$ if the test statistic is greater than $1.345$ .
The critical value is:
$\boxed{1.345}$

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Studdy Solution

STEP 1

1. The population variances are not equal.2. The sample sizes are 18 for County A and 8 for County B.3. The significance level α=0.10\alpha = 0.10α=0.10.4. The test is a one-tailed test since the claim is that the mean income in County A is greater than in County B.

STEP 2

1. Define the hypotheses.2. Determine the appropriate test statistic and distribution.3. Find the critical value(s) for the test.4. Identify the rejection region(s).

STEP 3

Define the hypotheses:- Null hypothesis (H0H_0H0​): μ1≤μ2\mu_1 \leq \mu_2μ1​≤μ2​ - Alternative hypothesis (HaH_aHa​): μ1>μ2\mu_1 > \mu_2μ1​>μ2​

STEP 4

Since the population variances are not equal and the sample sizes are small, use the Welch's t-test. The test statistic follows a t-distribution with degrees of freedom calculated using the Welch-Satterthwaite equation.

STEP 5

Find the critical value for a one-tailed test at α=0.10\alpha = 0.10α=0.10.- Use a t-distribution table or technology to find the critical value for the calculated degrees of freedom.Given the problem statement, the critical value is:tcritical=1.345 t_{critical} = 1.345 tcritical​=1.345

STEP 6

Identify the rejection region:- Since it is a one-tailed test, the rejection region is in the right tail of the t-distribution. - Reject H0H_0H0​ if the test statistic is greater than 1.3451.3451.345. The critical value is:1.345 \boxed{1.345} 1.345​

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. The population variances are not equal.
2. The sample sizes are 18 for County A and 8 for County B.
3. The significance level $\alpha = 0.10$ .
4. The test is a one-tailed test since the claim is that the mean income in County A is greater than in County B.

1. Define the hypotheses.
2. Determine the appropriate test statistic and distribution.
3. Find the critical value(s) for the test.
4. Identify the rejection region(s).

Define the hypotheses:
- Null hypothesis ( $H_0$ ): $\mu_1 \leq \mu_2$ - Alternative hypothesis ( $H_a$ ): $\mu_1 > \mu_2$

Find the critical value for a one-tailed test at $\alpha = 0.10$ .
- Use a t-distribution table or technology to find the critical value for the calculated degrees of freedom.
Given the problem statement, the critical value is:
$t_{critical} = 1.345$

Identify the rejection region:
- Since it is a one-tailed test, the rejection region is in the right tail of the t-distribution. - Reject $H_0$ if the test statistic is greater than $1.345$ .
The critical value is:
$\boxed{1.345}$