Answer :
To address the given hypothesis test, we will follow the specified steps:
### a. Compute the value of the test statistic
To compute the test statistic, we use the formula for the t-statistic in the context of a one-sample t-test:
[tex]\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \][/tex]
Given:
- Sample size [tex]\( n = 25 \)[/tex]
- Sample mean [tex]\( \bar{x} = 14 \)[/tex]
- Sample standard deviation [tex]\( s = 4.32 \)[/tex]
- Population mean under the null hypothesis [tex]\( \mu_0 = 12 \)[/tex]
Substituting in the given values, you get:
[tex]\[ t = \frac{14 - 12}{4.32 / \sqrt{25}} = \frac{2}{4.32 / 5} = \frac{2}{0.864} \approx 2.31 \][/tex]
Thus, the value of the test statistic is [tex]\( t = 2.31 \)[/tex].
### b. Compute the range for the [tex]\( p \)[/tex]-value
The [tex]\( p \)[/tex]-value is the probability of obtaining a test statistic at least as extreme as the one computed, assuming the null hypothesis is true. For a t-distribution with [tex]\( df = n - 1 = 25 - 1 = 24 \)[/tex] degrees of freedom, we look up the value 2.31 in a t-distribution table or use appropriate software to find the [tex]\( p \)[/tex]-value.
The computed [tex]\( p \)[/tex]-value is approximately [tex]\(\begin{cases} 0.015 \, \text{(rounded to 3 decimals)} \text{ for using software or tables} \end{cases}\)[/tex].
### c. Conclusion at [tex]\( \alpha = 0.05 \)[/tex]
To make a decision, compare the [tex]\( p \)[/tex]-value with the significance level [tex]\(\alpha\)[/tex]:
- If [tex]\( p \)[/tex]-value [tex]\( \leq \alpha \)[/tex], reject the null hypothesis.
- If [tex]\( p \)[/tex]-value [tex]\( > \alpha \)[/tex], do not reject the null hypothesis.
In our case, [tex]\( p \approx 0.015 \)[/tex] which is less than [tex]\(\alpha = 0.05\)[/tex].
Thus, we reject the null hypothesis. This conclusion indicates that there is sufficient evidence to support the claim that the population mean is greater than 12 at the 0.05 significance level.
### d. Rejection rule using the critical value
The critical value for a one-tailed t-test at [tex]\( \alpha = 0.05 \)[/tex] with 24 degrees of freedom can be found in the t-distribution table, which is approximately [tex]\( t_{0.05, 24} \approx 1.711 \)[/tex].
The rejection rule is:
- Reject [tex]\( H_0 \)[/tex] if the computed [tex]\( t \)[/tex]-statistic is greater than the critical value.
In our case:
- [tex]\( t = 2.31 \)[/tex] is indeed greater than the critical value [tex]\( 1.711 \)[/tex].
Thus, the rejection rule confirms our earlier decision. We can reject [tex]\( H_0 \)[/tex] and conclude that the population mean is greater than 12.
### Summary of Answers:
- a. The value of the test statistic [tex]\( t = 2.31 \)[/tex].
- b. The [tex]\( p \)[/tex]-value is approximately [tex]\( 0.015 \)[/tex].
- c. At [tex]\( \alpha = 0.05 \)[/tex], we conclude that the population mean is greater than 12. We reject the null hypothesis.
- d. The critical value is 1.711. We reject [tex]\( H_0 \)[/tex] if [tex]\( t \)[/tex] is greater than the critical value of 1.711. Yes, we can conclude that the population mean is greater than 12.
### a. Compute the value of the test statistic
To compute the test statistic, we use the formula for the t-statistic in the context of a one-sample t-test:
[tex]\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \][/tex]
Given:
- Sample size [tex]\( n = 25 \)[/tex]
- Sample mean [tex]\( \bar{x} = 14 \)[/tex]
- Sample standard deviation [tex]\( s = 4.32 \)[/tex]
- Population mean under the null hypothesis [tex]\( \mu_0 = 12 \)[/tex]
Substituting in the given values, you get:
[tex]\[ t = \frac{14 - 12}{4.32 / \sqrt{25}} = \frac{2}{4.32 / 5} = \frac{2}{0.864} \approx 2.31 \][/tex]
Thus, the value of the test statistic is [tex]\( t = 2.31 \)[/tex].
### b. Compute the range for the [tex]\( p \)[/tex]-value
The [tex]\( p \)[/tex]-value is the probability of obtaining a test statistic at least as extreme as the one computed, assuming the null hypothesis is true. For a t-distribution with [tex]\( df = n - 1 = 25 - 1 = 24 \)[/tex] degrees of freedom, we look up the value 2.31 in a t-distribution table or use appropriate software to find the [tex]\( p \)[/tex]-value.
The computed [tex]\( p \)[/tex]-value is approximately [tex]\(\begin{cases} 0.015 \, \text{(rounded to 3 decimals)} \text{ for using software or tables} \end{cases}\)[/tex].
### c. Conclusion at [tex]\( \alpha = 0.05 \)[/tex]
To make a decision, compare the [tex]\( p \)[/tex]-value with the significance level [tex]\(\alpha\)[/tex]:
- If [tex]\( p \)[/tex]-value [tex]\( \leq \alpha \)[/tex], reject the null hypothesis.
- If [tex]\( p \)[/tex]-value [tex]\( > \alpha \)[/tex], do not reject the null hypothesis.
In our case, [tex]\( p \approx 0.015 \)[/tex] which is less than [tex]\(\alpha = 0.05\)[/tex].
Thus, we reject the null hypothesis. This conclusion indicates that there is sufficient evidence to support the claim that the population mean is greater than 12 at the 0.05 significance level.
### d. Rejection rule using the critical value
The critical value for a one-tailed t-test at [tex]\( \alpha = 0.05 \)[/tex] with 24 degrees of freedom can be found in the t-distribution table, which is approximately [tex]\( t_{0.05, 24} \approx 1.711 \)[/tex].
The rejection rule is:
- Reject [tex]\( H_0 \)[/tex] if the computed [tex]\( t \)[/tex]-statistic is greater than the critical value.
In our case:
- [tex]\( t = 2.31 \)[/tex] is indeed greater than the critical value [tex]\( 1.711 \)[/tex].
Thus, the rejection rule confirms our earlier decision. We can reject [tex]\( H_0 \)[/tex] and conclude that the population mean is greater than 12.
### Summary of Answers:
- a. The value of the test statistic [tex]\( t = 2.31 \)[/tex].
- b. The [tex]\( p \)[/tex]-value is approximately [tex]\( 0.015 \)[/tex].
- c. At [tex]\( \alpha = 0.05 \)[/tex], we conclude that the population mean is greater than 12. We reject the null hypothesis.
- d. The critical value is 1.711. We reject [tex]\( H_0 \)[/tex] if [tex]\( t \)[/tex] is greater than the critical value of 1.711. Yes, we can conclude that the population mean is greater than 12.
