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

[tex]\[
\begin{array}{l}
H_o: \mu=78.2 \\
H_a: \mu\ \textless \ 78.2
\end{array}
\][/tex]

You believe the population is normally distributed and you know the standard deviation is [tex]\(\sigma=20.9\)[/tex]. You obtain a sample mean of [tex]\(\bar{x}=75.3\)[/tex] for a sample of size [tex]\(n=53\)[/tex].

1. What is the test statistic for this sample?
[tex]\[ \text{Test statistic} = \square \][/tex] (Report answer accurate to 3 decimal places.)

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

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

This test statistic leads to a decision to:
- reject the null
- accept the null
- fail to reject the null

Sagot :

GinnyAnsweridn
To solve this hypothesis testing problem, we need to follow several steps.

First, let's summarize the given information:
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(\mu = 78.2\)[/tex]
- Alternative hypothesis ([tex]\(H_a\)[/tex]): [tex]\(\mu < 78.2\)[/tex] (this is a left-tailed test)
- Significance level ([tex]\(\alpha\)[/tex]): 0.10
- Population standard deviation ([tex]\(\sigma\)[/tex]): 20.9
- Sample mean ([tex]\(\bar{x}\)[/tex]): 75.3
- Sample size ([tex]\(n\)[/tex]): 53

### Step 1: Calculate the Standard Error of the Mean

The standard error of the mean (SE) is calculated using the formula:
[tex]\[ SE = \frac{\sigma}{\sqrt{n}} \][/tex]
Substituting the known values:
[tex]\[ SE = \frac{20.9}{\sqrt{53}} \][/tex]

### Step 2: Calculate the Test Statistic (z-score)

The test statistic (z-score) for a single sample mean is calculated using the formula:
[tex]\[ z = \frac{\bar{x} - \mu}{SE} \][/tex]
Using the values:
[tex]\[ z = \frac{75.3 - 78.2}{SE} \][/tex]

### Step 3: Calculate the p-value

The p-value for a left-tailed test corresponds to the area under the standard normal curve to the left of the calculated z-score. This can be found using statistical tables or software tools.

### Step 4: Compare the p-value with [tex]\(\alpha\)[/tex] and make a decision

If the p-value is less than or equal to the significance level ([tex]\(\alpha = 0.10\)[/tex]), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

### Solution:

- The standard error is:
[tex]\[ SE = \frac{20.9}{\sqrt{53}} \approx 2.87 \][/tex]

- The test statistic (z-score) is:
[tex]\[ z = \frac{75.3 - 78.2}{2.87} \approx -1.01 \][/tex]

- The p-value for [tex]\( z = -1.01 \)[/tex] is approximately [tex]\( 0.1562 \)[/tex].

Comparing the p-value to the significance level:
- Since [tex]\( 0.1562 \)[/tex] is greater than [tex]\( \alpha = 0.10 \)[/tex], we fail to reject the null hypothesis.

### Final Answers:

1. Test Statistic [tex]\( z \)[/tex] = [tex]\(-1.01\)[/tex] (accurate to 3 decimal places)
2. p-value = [tex]\( 0.1562 \)[/tex] (accurate to 4 decimal places)
3. The p-value is greater than [tex]\(\alpha\)[/tex].
4. The test statistic leads to a decision to fail to reject the null hypothesis.
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