To test the null hypothesis [tex]\( H_0: \mu_0 = 40 \)[/tex] against the alternative hypothesis [tex]\( H_1: \mu \neq 40 \)[/tex], we follow these steps:
1. Sample Information:
- Sample size [tex]\( n = 25 \)[/tex]
- Population standard deviation [tex]\( \sigma = 6 \)[/tex]
- Sample mean [tex]\( \bar{x} = 42.3 \)[/tex]
- Null hypothesis mean [tex]\( \mu_0 = 40 \)[/tex]
2. Test Statistic Calculation (z-score):
We calculate the z-score using the formula:
[tex]\[
z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}
\][/tex]
Plugging in the values:
[tex]\[
z = \frac{42.3 - 40}{6 / \sqrt{25}} = \frac{2.3}{1.2} = 1.9167
\][/tex]
3. P-value Calculation for a Two-Tailed Test:
To find the P-value, we determine the probability that the test statistic is as extreme as or more extreme than the observed value under the null hypothesis in both tails of the normal distribution.
For a z-score of 1.9167, we look up this value in the standard normal distribution table or use a calculator to find the cumulative probability.
[tex]\[
P(Z \leq 1.9167) \approx 0.9724
\][/tex]
Since this is a two-tailed test, we need to consider both tails. Hence, we double the cumulative probability of the right tail beyond the absolute z-value:
[tex]\[
\text{P-value} = 2 \times (1 - 0.9724) = 2 \times 0.0276 = 0.0553
\][/tex]
Therefore, the P-value is approximately [tex]\( 0.0553 \)[/tex].
Thus, the correct answer to the question is:
[tex]\[
\boxed{\text{P-value is about 0.0553}}
\][/tex]
