To solve this problem step-by-step:
1. State the hypotheses:
- Null hypothesis ([tex]\( H_0 \)[/tex]): [tex]\(\mu = 125500\)[/tex]
- Alternative hypothesis ([tex]\( H_a \)[/tex]): [tex]\(\mu \neq 125500\)[/tex]
2. Given data:
- Population mean ([tex]\(\mu_0\)[/tex]): [tex]\( 125500 \)[/tex]
- Sample mean ([tex]\(\bar{x}\)[/tex]): [tex]\( 126584.4 \)[/tex]
- Population standard deviation ([tex]\(\sigma\)[/tex]): [tex]\( 4000 \)[/tex]
- Sample size ([tex]\( n \)[/tex]): [tex]\( 60 \)[/tex]
- Significance level ([tex]\(\alpha\)[/tex]): [tex]\( 0.05 \)[/tex]
3. Calculate the standard error of the mean:
[tex]\[
\text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}} = \frac{4000}{\sqrt{60}} = 516.3978
\][/tex]
4. Calculate the test statistic (z-score):
[tex]\[
z = \frac{\bar{x} - \mu_0}{\text{SE}} = \frac{126584.4 - 125500}{516.3978} = 2.0999
\][/tex]
Reported to 3 decimal places: [tex]\( z = 2.100 \)[/tex]
5. Find the p-value for the calculated z-score in a two-tailed test:
[tex]\[
\text{p-value} = 2 \times \left(1 - \Phi(|z|)\right)
\][/tex]
where [tex]\(\Phi\)[/tex] is the cumulative distribution function (CDF) of the standard normal distribution.
Plugging in the value:
[tex]\[
\text{p-value} = 2 \times (1 - \Phi(2.0999)) \approx 0.0357
\][/tex]
6. Compare the p-value with the significance level:
- The p-value [tex]\( \approx 0.0357 \)[/tex]
- Since [tex]\( 0.0357 < 0.05 \)[/tex]
Therefore, the required answers are:
- The test statistic for this sample: [tex]\( \boxed{2.100} \)[/tex]
- The p-value for this sample: [tex]\( \boxed{0.0357} \)[/tex]
- The p-value is less than (or equal to) [tex]\(\alpha\)[/tex]
