To test [tex]\( H_0: \mu = 40 \)[/tex] versus [tex]\( H_1: \mu \neq 40 \)[/tex], we are given the following data:
- Population mean under null hypothesis ([tex]\( \mu_0 \)[/tex]): 40
- Sample size ([tex]\( n \)[/tex]): 25
- Population standard deviation ([tex]\( \sigma \)[/tex]): 6
- Sample mean ([tex]\( \bar{x} \)[/tex]): 42.3
- Significance level ([tex]\( \alpha \)[/tex]): 0.01
We need to perform a hypothesis test using a TI 83/84 calculator. Here is the step-by-step solution:
1. Calculate the z-score:
[tex]\[
z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}
\][/tex]
Substituting in the given values:
[tex]\[
z = \frac{42.3 - 40}{6 / \sqrt{25}} = \frac{2.3}{6 / 5} = \frac{2.3}{1.2} \approx 1.92
\][/tex]
2. Calculate the P-value for a two-tailed test:
Using the z-score table or the use of a calculator, find the area under the standard normal curve corresponding to the computed z-score.
The P-value for [tex]\( z \approx 1.92 \)[/tex] is:
[tex]\[
P(\text{value}) \approx 0.055
\][/tex]
3. Decision Rule:
Compare the P-value with the significance level [tex]\(\alpha\)[/tex]:
- If [tex]\( \text{P-value} < \alpha \)[/tex], reject the null hypothesis.
- If [tex]\( \text{P-value} \geq \alpha \)[/tex], do not reject the null hypothesis.
Given:
[tex]\[
\text{P-value} \approx 0.055 \quad \text{and} \quad \alpha = 0.01
\][/tex]
Since [tex]\( 0.055 > 0.01 \)[/tex],
Decision: Accept the null hypothesis.
Reason: The P-value is greater than the significance level. Therefore, there is not enough evidence to reject the null hypothesis.
4. Conclusion:
We conclude that, at the [tex]\( \alpha = 0.01 \)[/tex] significance level, the sample data does not provide sufficient evidence to reject the null hypothesis. Thus, we accept [tex]\( H_0: \mu = 40 \)[/tex].
So the final correct interpretation from the options given would be:
- My P-value greater than a Alpha, so I Accept Null Hypothesis
