Answer :
To solve this problem, we need to conduct a hypothesis test to determine whether the sample provides enough evidence to reject the null hypothesis in favor of the alternative hypothesis.
### Step 1: Identify the Hypotheses
Given hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu = 87.2\)[/tex]
- Alternative hypothesis ([tex]\(H_1\)[/tex]): [tex]\(\mu > 87.2\)[/tex]
### Step 2: Determine the Test Statistic
Since the population standard deviation is unknown and the sample size is small ([tex]\(n = 26\)[/tex]), we use the t-test statistic. The formula for the t-test statistic is:
[tex]\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\][/tex]
Where:
- [tex]\(\bar{x} = 97.6\)[/tex] (sample mean)
- [tex]\(\mu_0 = 87.2\)[/tex] (hypothesized mean)
- [tex]\(s = 15.2\)[/tex] (sample standard deviation)
- [tex]\(n = 26\)[/tex] (sample size)
Plug in the values:
1. Calculate the standard error [tex]\(SE = \frac{s}{\sqrt{n}}\)[/tex].
2. Calculate the test statistic:
[tex]\[
t = \frac{97.6 - 87.2}{15.2 / \sqrt{26}} \approx 3.489
\][/tex]
### Step 3: Find the P-value
The p-value is the probability that the test statistic would be as extreme as, or more extreme than, the value calculated, under the null hypothesis. For this t-test with [tex]\(n - 1 = 25\)[/tex] degrees of freedom:
Calculate the p-value using the cumulative distribution function (CDF) of the t-distribution:
[tex]\[
\text{p-value} = 1 - \text{CDF}(t, df = 25) \approx 0.0009
\][/tex]
### Step 4: Compare the P-value with the Significance Level
The significance level [tex]\(\alpha = 0.01\)[/tex].
Since the p-value [tex]\((0.0009)\)[/tex] is less than [tex]\(\alpha = 0.01\)[/tex], we reject the null hypothesis [tex]\(H_0\)[/tex].
### Conclusion
There is sufficient evidence to support the claim that the population mean [tex]\(\mu\)[/tex] is greater than 87.2 at the 0.01 significance level.
### Step 1: Identify the Hypotheses
Given hypotheses:
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu = 87.2\)[/tex]
- Alternative hypothesis ([tex]\(H_1\)[/tex]): [tex]\(\mu > 87.2\)[/tex]
### Step 2: Determine the Test Statistic
Since the population standard deviation is unknown and the sample size is small ([tex]\(n = 26\)[/tex]), we use the t-test statistic. The formula for the t-test statistic is:
[tex]\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\][/tex]
Where:
- [tex]\(\bar{x} = 97.6\)[/tex] (sample mean)
- [tex]\(\mu_0 = 87.2\)[/tex] (hypothesized mean)
- [tex]\(s = 15.2\)[/tex] (sample standard deviation)
- [tex]\(n = 26\)[/tex] (sample size)
Plug in the values:
1. Calculate the standard error [tex]\(SE = \frac{s}{\sqrt{n}}\)[/tex].
2. Calculate the test statistic:
[tex]\[
t = \frac{97.6 - 87.2}{15.2 / \sqrt{26}} \approx 3.489
\][/tex]
### Step 3: Find the P-value
The p-value is the probability that the test statistic would be as extreme as, or more extreme than, the value calculated, under the null hypothesis. For this t-test with [tex]\(n - 1 = 25\)[/tex] degrees of freedom:
Calculate the p-value using the cumulative distribution function (CDF) of the t-distribution:
[tex]\[
\text{p-value} = 1 - \text{CDF}(t, df = 25) \approx 0.0009
\][/tex]
### Step 4: Compare the P-value with the Significance Level
The significance level [tex]\(\alpha = 0.01\)[/tex].
Since the p-value [tex]\((0.0009)\)[/tex] is less than [tex]\(\alpha = 0.01\)[/tex], we reject the null hypothesis [tex]\(H_0\)[/tex].
### Conclusion
There is sufficient evidence to support the claim that the population mean [tex]\(\mu\)[/tex] is greater than 87.2 at the 0.01 significance level.