College

You wish to test the following claim [tex](H_a)[/tex] at a significance level of [tex]\alpha=0.005[/tex].

\[
\begin{array}{l}
H_0: \mu = 75.2 \\
H_a: \mu \neq 75.2
\end{array}
\]

You believe the population is normally distributed, but you do not know the standard deviation. You obtain the following sample of data:

\[
\begin{array}{|r|r|r|r|r|}
\hline
63 & 71.4 & 66.4 & 64 & 66.6 \\
\hline
69.7 & 80 & 76.7 & 73.3 & 66.6 \\
\hline
66.6 & 70.3 & 72.4 & 78 & 58.2 \\
\hline
69.7 & 70.5 & 72.2 & 73.3 & 83.8 \\
\hline
66.1 & 65.3 & 71.2 & 65 & 68.1 \\
\hline
71.6 & 72.4 & 65.3 & 79.4 & 66.9 \\
\hline
64 & 76 & 66.9 & 57.3 & 68.3 \\
\hline
60.2 & 75.3 & 75.3 & 68.8 & 65.5 \\
\hline
83.3 & 59 & 66.9 & 86.7 & 66.6 \\
\hline
89.3 & 81.5 & 76.4 & 72.9 & 61.7 \\
\hline
82.8 & 66.4 & 73.9 & 57.3 & 69.9 \\
\hline
\end{array}
\]

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
Test statistic [tex]= \square[/tex]

What is the [tex]p[/tex]-value for this sample? (Report answer accurate to four decimal places.)
[tex]p[/tex]-value [tex]= \square[/tex]

Answer :

To find the test statistic and the p-value for the given hypothesis test, we'll follow these steps:

### Step 1: Understand the Setup

We are conducting a hypothesis test for the population mean [tex]\(\mu\)[/tex] with the following hypotheses:
- Null Hypothesis [tex]\(H_0: \mu = 75.2\)[/tex]
- Alternative Hypothesis [tex]\(H_a: \mu \neq 75.2\)[/tex]

The significance level is [tex]\(\alpha = 0.005\)[/tex].

### Step 2: Collect and Analyze the Sample Data

You have a sample data set with 55 observations. Given that the population standard deviation is unknown and the sample size is relatively large but not enormous, we will use the t-distribution.

### Step 3: Calculate the Sample Mean and Sample Standard Deviation

1. Sample Mean ([tex]\(\bar{x}\)[/tex]): Add all the sample values together and divide by the number of observations to find the sample mean.

2. Sample Standard Deviation (s): Use the formula for standard deviation with Bessel's correction (using [tex]\(n-1\)[/tex] as the denominator).

### Step 4: Calculate the Test Statistic

The test statistic for the mean when the population standard deviation is unknown is a t-statistic, given by the formula:

[tex]\[
t = \frac{\bar{x} - \mu}{s / \sqrt{n}}
\][/tex]

Where:
- [tex]\(\bar{x}\)[/tex] is the sample mean
- [tex]\(\mu\)[/tex] is the population mean under the null hypothesis ([tex]\(75.2\)[/tex])
- [tex]\(s\)[/tex] is the sample standard deviation
- [tex]\(n\)[/tex] is the sample size

After performing the calculation, the test statistic is found to be:

[tex]\[
t \approx -4.794
\][/tex]

### Step 5: Calculate the p-value

Given that the alternative hypothesis is two-tailed ([tex]\(H_a: \mu \neq 75.2\)[/tex]), the p-value is calculated by finding the probability that a t-distribution with [tex]\(n-1\)[/tex] degrees of freedom is more extreme than the observed test statistic in either tail.

Using statistical software or a t-distribution table, you find the p-value to be:

[tex]\[
\text{p-value} \approx 0.0000
\][/tex]

### Conclusion

Finally, with a test statistic of approximately [tex]\(-4.794\)[/tex] and a p-value of approximately [tex]\(0.0000\)[/tex], you compare the p-value to the significance level [tex]\(\alpha = 0.005\)[/tex]. Since the p-value is smaller than [tex]\(\alpha\)[/tex], you reject the null hypothesis [tex]\(H_0\)[/tex]. This suggests there is significant evidence to support the claim that the mean is different from 75.2.