Question

You wish to test the following claim (H_a) at a significance level of alpha= 0.10.

H_o : μ = 83.9
H_a : μ < 83.9

You believe the population is normally distributed and you know the standard deviation is σ=14.8. You obtain a sample mean of M = 81 for a sample of size n = 57.

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic = _____


What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value = _____


The p-value is... ___
A. less than (or equal to)
B. greater than


This test statistic leads to a decision to... ___
A. reject the null
B. accept the null
C. fail to reject the null

Answer 1

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**