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Suppose Wall Street securities firms paid out year-end bonuses of [tex]\$125,500[/tex] per employee last year. We take a sample of employees at the ASBE securities firm to see whether the mean year-end bonus is different from the reported mean of [tex]\$125,500[/tex] for the population.

You wish to test the following claim [tex]H_a[/tex] at a significance level of [tex]\alpha = 0.05[/tex].

[tex]\[
\begin{array}{l}
H_0: \mu = 125500 \\
H_a: \mu \neq 125500
\end{array}
\][/tex]

You believe the population is normally distributed and you know the standard deviation is [tex]\sigma = 4000[/tex]. You obtain a sample mean of [tex]\bar{x} = 126584.4[/tex] for a sample of size [tex]n = 60[/tex].

1. What is the test statistic for this sample?
Test statistic [tex]= \square[/tex] (Report answer accurate to 3 decimal places.)

2. What is the p-value for this sample?
p-value [tex]= \square[/tex] (Report answer accurate to 4 decimal places.)

The p-value is...
A. less than (or equal to) [tex]\alpha[/tex]
B. greater than [tex]\alpha[/tex]



Answer :

GinnyAnswer
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]