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You wish to test the following claim [tex]\(\left(H_a\right)\)[/tex] at a significance level of [tex]\(\alpha=0.001\)[/tex].

[tex]\[
\begin{array}{l}
H_o: \mu = 65.8 \\
H_a: \mu \ \textless \ 65.8
\end{array}
\][/tex]

You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size [tex]\(n = 14\)[/tex] with a mean [tex]\(\bar{x} = 61.9\)[/tex] and a standard deviation of [tex]\(s = 9.2\)[/tex].

a. What is the test statistic for this sample?
[tex]\[
\text{Test statistic} = \boxed{\hspace{2cm}} \quad \text{(Round to 3 decimal places)}
\][/tex]

b. What is the p-value for this sample?
[tex]\[
\text{p-value} = \boxed{\hspace{2cm}} \quad \text{(Use technology. Round to 4 decimal places)}
\][/tex]

c. The p-value is...
[tex]\[
\begin{array}{l}
\boxed{\text{less than (or equal to) } \alpha} \\
\boxed{\text{greater than } \alpha}
\end{array}
\][/tex]

d. This test statistic leads to a decision to...
[tex]\[
\begin{array}{l}
\boxed{\text{reject the null}} \\
\boxed{\text{accept the null}} \\
\boxed{\text{fail to reject the null}}
\end{array}
\][/tex]

Sagot :

GinnyAnsweridn
Let's proceed step-by-step through the hypothesis testing process.

### Step 1: Define the Hypotheses
Given hypotheses are:
[tex]\[ H_0: \mu = 65.8 \\ H_a: \mu < 65.8 \][/tex]
Here, we're conducting a one-tailed test to see if the sample mean is significantly less than 65.8.

### Step 2: Gather Given Information
- Alpha ([tex]\(\alpha\)[/tex]) = 0.001
- Population mean under the null hypothesis ([tex]\(\mu_0\)[/tex]) = 65.8
- Sample size ([tex]\(n\)[/tex]) = 14
- Sample mean ([tex]\(\bar{x}\)[/tex]) = 61.9
- Sample standard deviation ([tex]\(s\)[/tex]) = 9.2

### Step 3: Calculate the Test Statistic
We use the t-statistic for hypothesis testing when the population standard deviation is unknown:
[tex]\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \][/tex]
Substituting in the given values:
[tex]\[ t = \frac{61.9 - 65.8}{9.2 / \sqrt{14}} = \frac{-3.9}{2.459} \approx -1.586 \][/tex]

Thus, the test statistic is:
[tex]\[ \text{test statistic} = -1.586 \][/tex]

### Step 4: Calculate the P-Value
The p-value in a one-tailed t-test can be found using the cumulative distribution function (CDF) for the t-distribution with [tex]\( n - 1 \)[/tex] degrees of freedom:
[tex]\[ \text{degrees of freedom} = 14 - 1 = 13 \][/tex]
Using the CDF of the t-distribution:
[tex]\[ \text{p-value} \approx 0.0684 \][/tex]

Therefore, the p-value is:
[tex]\[ \text{p-value} = 0.0684 \][/tex]

### Step 5: Compare the P-Value with Alpha
Next, we compare the p-value with the significance level [tex]\(\alpha = 0.001\)[/tex]:
[tex]\[ 0.0684 > 0.001 \][/tex]

So, the p-value is:
[tex]\[ \text{greater than } \alpha \][/tex]

### Step 6: Make the Decision
Based on the comparison of the p-value with [tex]\(\alpha\)[/tex]:
- If the p-value is less than or equal to [tex]\(\alpha\)[/tex], we reject the null hypothesis.
- If the p-value is greater than [tex]\(\alpha\)[/tex], we fail to reject the null hypothesis.

Since [tex]\(0.0684\)[/tex] is greater than [tex]\(0.001\)[/tex], we:
[tex]\[ \text{fail to reject the null} \][/tex]

### Summary
a. The test statistic for this sample is:
[tex]\[ \text{test statistic} = -1.586 \][/tex]

b. The p-value for this sample is:
[tex]\[ \text{p-value} = 0.0684 \][/tex]

c. The p-value is:
[tex]\[ \text{greater than } \alpha \][/tex]

d. This test statistic leads to a decision to:
[tex]\[ \text{fail to reject the null} \][/tex]
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