Answered

You wish to test the following claim [tex]\left(H_a\right)[/tex] at a significance level of [tex]\alpha=0.10[/tex].

[tex]\[
\begin{array}{l}
H_o: \mu=66.3 \\
H_a: \mu\ \textgreater \ 66.3
\end{array}
\][/tex]

You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size [tex]n=84[/tex] with mean [tex]\bar{x}=68[/tex] and a standard deviation of [tex]s=8.8[/tex].

1. What is the test statistic for this sample?

Test statistic [tex]= \square[/tex] (Report answer accurate to 3 decimal places.)

2. What is the [tex]p[/tex]-value for this sample?

[tex]p[/tex]-value [tex]= \square[/tex] (Report answer accurate to 4 decimal places.)

3. The [tex]p[/tex]-value is...

- less than (or equal to) [tex]\alpha[/tex]
- greater than [tex]\alpha[/tex]

4. This test statistic leads to a decision to...

- reject the null
- accept the null
- fail to reject the null



Answer :

GinnyAnswer
To test the claim at a significance level of [tex]\(\alpha = 0.10\)[/tex], we need to perform a one-sample t-test. Here is a detailed, step-by-step solution:

### Step 1: State the Hypotheses
The null hypothesis ([tex]\(H_0\)[/tex]) and the alternative hypothesis ([tex]\(H_a\)[/tex]) are given by:
[tex]\[ H_0: \mu = 66.3 \\ H_a: \mu > 66.3 \][/tex]

### Step 2: Gather the Sample Information
The sample information provided is:
- Sample size ([tex]\(n\)[/tex]) = 84
- Sample mean ([tex]\(\bar{x}\)[/tex]) = 68
- Sample standard deviation ([tex]\(s\)[/tex]) = 8.8

### Step 3: Calculate the Test Statistic
The test statistic for a one-sample t-test is calculated using the formula:
[tex]\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \][/tex]
where:
- [tex]\(\bar{x}\)[/tex] is the sample mean
- [tex]\(\mu_0\)[/tex] is the population mean under the null hypothesis
- [tex]\(s\)[/tex] is the sample standard deviation
- [tex]\(n\)[/tex] is the sample size

Plugging in the given values:
[tex]\[ t = \frac{68 - 66.3}{8.8 / \sqrt{84}} \][/tex]

Performing the calculation gives:
[tex]\[ t \approx 1.771 \][/tex]

### Step 4: Determine the Degrees of Freedom
The degrees of freedom ([tex]\(df\)[/tex]) for this test is:
[tex]\[ df = n - 1 = 84 - 1 = 83 \][/tex]

### Step 5: Calculate the p-value
The p-value for this test is obtained by finding the area to the right of the calculated test statistic [tex]\(t\)[/tex] in the t-distribution with 83 degrees of freedom. This corresponds to:
[tex]\[ p = P(T > 1.771) \][/tex]

The calculated p-value is:
[tex]\[ p \approx 0.0402 \][/tex]

### Step 6: Make a Decision
Compare the p-value to the significance level [tex]\(\alpha\)[/tex]:
[tex]\[ p = 0.0402 \quad \text{and} \quad \alpha = 0.10 \][/tex]

Since the p-value is less than [tex]\(\alpha\)[/tex], we have:
[tex]\[ \text{p-value } \leq \alpha \][/tex]

### Conclusion
Based on the p-value, we make the following decision:
[tex]\[ \text{Decision: } \text{reject the null hypothesis} \][/tex]

### Summary of Results
- The test statistic is: [tex]\( t = 1.771 \)[/tex]
- The p-value is: [tex]\( p = 0.0402 \)[/tex]
- The p-value is less than or equal to [tex]\(\alpha\)[/tex]: (yes, [tex]\(0.0402 \leq 0.10\)[/tex])
- Therefore, we reject the null hypothesis.

Thus, the final answers are:

1. Test statistic = 1.771
2. p-value = 0.0402
3. The p-value is less than (or equal to) [tex]\(\alpha\)[/tex]
4. This test statistic leads to a decision to reject the null hypothesis.