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Sagot :
Sure, let's go through this step by step.
a. Compute the value of the test statistic:
Given:
- Sample mean, [tex]\(\bar{x} = 14\)[/tex]
- Population mean under null hypothesis, [tex]\(\mu = 12\)[/tex]
- Sample standard deviation, [tex]\(s = 4.3\)[/tex]
- Sample size, [tex]\(n = 25\)[/tex]
To compute the test statistic for a one-sample t-test, we use the formula:
[tex]\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \][/tex]
Substitute the given values into the formula:
[tex]\[ t = \frac{14 - 12}{4.3 / \sqrt{25}} = \frac{2}{4.3 / 5} = \frac{2}{0.86} \approx 2.31 \][/tex]
b. Compute a range for the p-value:
We need to use the t-distribution with [tex]\(df = n - 1 = 25 - 1 = 24\)[/tex] degrees of freedom. Given the test statistic [tex]\(t = 2.31\)[/tex], we refer to the t-distribution table to find the range for the p-value.
From the table (or the given result), we know:
[tex]\[ 0.01 < p \leq 0.025 \][/tex]
Thus, the [tex]\( p\)[/tex]-value is between 0.01 and 0.025.
c. Using [tex]\(\alpha=0.05\)[/tex], what is the critical value for the test statistic?
For a one-tailed test at the 0.05 significance level with [tex]\(df = 24\)[/tex]:
Using the t-distribution table or a critical value calculator:
[tex]\[ \text{Critical value} = 1.711 \][/tex]
Rejection rule: Reject [tex]\(H_0\)[/tex] if [tex]\(t\)[/tex] is greater than or equal to the critical value.
[tex]\[ \text{Reject } H_0 \text{ if } t \geq 1.711 \][/tex]
d. Using [tex]\(\alpha = 0.01\)[/tex], can you conclude that the population mean is greater than 12?
For a one-tailed test at the 0.01 significance level with [tex]\(df = 24\)[/tex]:
Using the t-distribution table or a critical value calculator:
[tex]\[ \text{Critical value} = 2.492 \][/tex]
Compare the test statistic [tex]\(t = 2.31\)[/tex] with the critical value [tex]\(2.492\)[/tex]:
Since [tex]\(2.31 < 2.492\)[/tex], we fail to reject the null hypothesis at the 0.01 significance level.
Therefore, the answer is:
[tex]\[ \text{No} \][/tex]
Summarizing the steps:
- a. Test statistic value: [tex]\(2.31\)[/tex]
- b. [tex]\(p\)[/tex]-value range: between 0.01 and 0.025
- c. Critical value for [tex]\(\alpha = 0.05\)[/tex]: [tex]\(1.711\)[/tex]
- Rejection rule: Reject [tex]\(H_0\)[/tex] if [tex]\(t \geq 1.711\)[/tex]
- d. At [tex]\(\alpha = 0.01\)[/tex], we cannot conclude that the population mean is greater than 12: [tex]\(\text{No}\)[/tex]
a. Compute the value of the test statistic:
Given:
- Sample mean, [tex]\(\bar{x} = 14\)[/tex]
- Population mean under null hypothesis, [tex]\(\mu = 12\)[/tex]
- Sample standard deviation, [tex]\(s = 4.3\)[/tex]
- Sample size, [tex]\(n = 25\)[/tex]
To compute the test statistic for a one-sample t-test, we use the formula:
[tex]\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \][/tex]
Substitute the given values into the formula:
[tex]\[ t = \frac{14 - 12}{4.3 / \sqrt{25}} = \frac{2}{4.3 / 5} = \frac{2}{0.86} \approx 2.31 \][/tex]
b. Compute a range for the p-value:
We need to use the t-distribution with [tex]\(df = n - 1 = 25 - 1 = 24\)[/tex] degrees of freedom. Given the test statistic [tex]\(t = 2.31\)[/tex], we refer to the t-distribution table to find the range for the p-value.
From the table (or the given result), we know:
[tex]\[ 0.01 < p \leq 0.025 \][/tex]
Thus, the [tex]\( p\)[/tex]-value is between 0.01 and 0.025.
c. Using [tex]\(\alpha=0.05\)[/tex], what is the critical value for the test statistic?
For a one-tailed test at the 0.05 significance level with [tex]\(df = 24\)[/tex]:
Using the t-distribution table or a critical value calculator:
[tex]\[ \text{Critical value} = 1.711 \][/tex]
Rejection rule: Reject [tex]\(H_0\)[/tex] if [tex]\(t\)[/tex] is greater than or equal to the critical value.
[tex]\[ \text{Reject } H_0 \text{ if } t \geq 1.711 \][/tex]
d. Using [tex]\(\alpha = 0.01\)[/tex], can you conclude that the population mean is greater than 12?
For a one-tailed test at the 0.01 significance level with [tex]\(df = 24\)[/tex]:
Using the t-distribution table or a critical value calculator:
[tex]\[ \text{Critical value} = 2.492 \][/tex]
Compare the test statistic [tex]\(t = 2.31\)[/tex] with the critical value [tex]\(2.492\)[/tex]:
Since [tex]\(2.31 < 2.492\)[/tex], we fail to reject the null hypothesis at the 0.01 significance level.
Therefore, the answer is:
[tex]\[ \text{No} \][/tex]
Summarizing the steps:
- a. Test statistic value: [tex]\(2.31\)[/tex]
- b. [tex]\(p\)[/tex]-value range: between 0.01 and 0.025
- c. Critical value for [tex]\(\alpha = 0.05\)[/tex]: [tex]\(1.711\)[/tex]
- Rejection rule: Reject [tex]\(H_0\)[/tex] if [tex]\(t \geq 1.711\)[/tex]
- d. At [tex]\(\alpha = 0.01\)[/tex], we cannot conclude that the population mean is greater than 12: [tex]\(\text{No}\)[/tex]
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