To find the test statistic for the given sample, we can use the formula for the z-test:
\[ z = \frac{M - \mu_0}{\sigma / \sqrt{n}} \]
where:
- \( M \) is the sample mean,
- \( \mu_0 \) is the population mean under the null hypothesis,
- \( \sigma \) is the population standard deviation,
- \( n \) is the sample size.
Substituting the values:
\[ M = 81, \quad \mu_0 = 83.9, \quad \sigma = 14.8, \quad n = 57 \]
Calculating the standard error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{14.8}{\sqrt{57}} \approx \frac{14.8}{7.550} \approx 1.961 \]
Now calculate the z-test statistic:
\[ z = \frac{81 - 83.9}{1.961} \approx \frac{-2.9}{1.961} \approx -1.477 \]
So the test statistic is:
\[ \text{test statistic} = -1.477 \quad (\text{accurate to three decimal places: } -1.477) \]
Next, we need to find the p-value associated with the test statistic. The p-value represents the probability of observing a test statistic as extreme as (or more extreme than) the observed value under the null hypothesis. Since we are conducting a left-tailed test, we can look up the area to the left of \( z = -1.477 \).
Using the standard normal distribution table or a calculator, we find:
\[ p\text{-value} \approx 0.0703 \quad (\text{accurate to four decimal places: } 0.0703) \]
This means the p-value is:
\[ \text{p-value} = 0.0703 \]
Next, we compare the p-value to the significance level (\( \alpha = 0.10 \)):
\[ 0.0703 < 0.10 \]
Thus, the p-value is:
The p-value is... A. less than (or equal to)
This test statistic leads to a decision to... A. reject the null because the p-value is less than the significance level.
So summarizing the answers:
- Test statistic: -1.477
- p-value: 0.0703
- The p-value is... A. less than (or equal to)
- This test statistic leads to a decision to... A. reject the null
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