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

[tex]\[
\begin{array}{l}
H_o: \mu=71.2 \\
H_a: \mu\ \textless \ 71.2
\end{array}
\][/tex]

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

[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline
Column A & Column B & Column C & Column D & Column E \\
\hline
48.3 & 60.9 & 78.9 & 82.6 & 55.7 \\
64.3 & 70.8 & 66.9 & 65.2 & 59.6 \\
60.3 & 62.2 & 53.3 & 68.1 & 77.4 \\
66.9 & 63.1 & 49.2 & 56.2 & 72.1 \\
74.5 & 90.2 & 73.5 & 63.1 & 68.4 \\
78.4 & 47.3 & 62.8 & 51.5 & 55.3 \\
72.1 & 79.4 & 61.9 & 64.0 & 58.2 \\
62.5 & 61.6 & 58.5 & 59.6 & 69.0 \\
66.4 & 49.2 & 56.2 & 64.9 & 77.9 \\
66.1 & 70.2 & 70.5 & 77.0 & 67.2 \\
73.5 & 65.2 & 71.8 & 66.9 & 69.3 \\
\hline
\end{tabular}
\][/tex]

Note: To save vertical scrolling space, the data set is shown here with five columns. In Statcrunch, you will need to make sure all the data are in ONE column.

Use technology.

1. What is the test statistic for this sample?
Test statistic [tex]\( = \)[/tex] [tex]\(\square\)[/tex] (Report answer accurate to 4 decimal places.)

2. What is the p-value for this sample?
p-value [tex]\( = \)[/tex] [tex]\(\square\)[/tex] (Report answer accurate to 4 decimal places.)

The p-value is...
- less than (or equal to) [tex]\(\alpha\)[/tex]
- greater than [tex]\(\alpha\)[/tex]



Answer :

To address the hypothesis test, let's break down each step involved.

### 1. Formulating the Hypotheses
The null and alternative hypotheses are given as:
- [tex]\( H_0: \mu = 71.2 \)[/tex]
- [tex]\( H_a: \mu < 71.2 \)[/tex]

where [tex]\(\mu\)[/tex] represents the population mean.

### 2. Significance Level
The significance level ([tex]\(\alpha\)[/tex]) is specified as 0.005.

### 3. Data Collection
We are provided with the following array of data points:

[tex]\[ \begin{array}{|r|r|r|r|r|} \hline 48.3 & 60.9 & 78.9 & 82.6 & 55.7 \\ 64.3 & 70.8 & 66.9 & 65.2 & 59.6 \\ 60.3 & 62.2 & 53.3 & 68.1 & 77.4 \\ 66.9 & 63.1 & 49.2 & 56.2 & 72.1 \\ 74.5 & 90.2 & 73.5 & 63.1 & 68.4 \\ 78.4 & 47.3 & 62.8 & 51.5 & 55.3 \\ 72.1 & 79.4 & 61.9 & 64 & 58.2 \\ 62.5 & 61.6 & 58.5 & 59.6 & 69 \\ 66.4 & 49.2 & 56.2 & 64.9 & 77.9 \\ 66.1 & 70.2 & 70.5 & 77 & 67.2 \\ 73.5 & 65.2 & 71.8 & 66.9 & 69.3 \\ \hline \end{array} \][/tex]

### 4. Calculations
#### A. Sample Size
The sample size ([tex]\(n\)[/tex]) is 55.

#### B. Sample Mean ([tex]\(\bar{x}\)[/tex])
The mean of the sample data is calculated to be approximately 64.7818.

#### C. Sample Standard Deviation (s)
The sample standard deviation is approximately 8.1565.

#### D. Test Statistic (t)
The test statistic is calculated using the formula:
[tex]\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \][/tex]
Substituting the values:
[tex]\[ t = \frac{64.7818 - 71.2}{\frac{8.1565}{\sqrt{55}}} \approx -4.5778 \][/tex]

#### E. Degrees of Freedom
The degrees of freedom (df) is [tex]\(n - 1 = 54\)[/tex].

### 5. P-value Calculation
The P-value for the one-tailed test is found using the t-distribution with 54 degrees of freedom at the obtained t-statistic:
[tex]\[ \text{P-value} \approx 0.0 \][/tex]

### 6. Conclusion
Since the P-value (0.0) is less than the significance level (0.005), we reject the null hypothesis.

### Summary
- Test Statistic: [tex]\( t = -4.5778 \)[/tex]
- P-value: [tex]\( \text{P-value} = 0.0 \)[/tex]
- Conclusion: The P-value is less than or equal to [tex]\(\alpha\)[/tex].

The test statistic [tex]\(-4.5778\)[/tex] is accurately reported to 4 decimal places. The p-value is [tex]\(0.0\)[/tex], indicating strong evidence against the null hypothesis. Hence, we conclude that the population mean is likely less than 71.2 at the 0.005 significance level.