Question

The mean hourly wage for employees in industries is currently $24.57. Suppose we take a sample of employees from the manufacturing industry to see if the mean hourly wage differs from the reported mean of $24.57 for the U.S industries. a. State the null and alternative hypotheses we should use to test whether the population mean hourly wage in the manufacturing industry differs from the population mean hourly wage in the U.S industries. b. Suppose a sample of 30 employees from the manufacturing industry showed a sample mean of $23.89 per hour. Assume a population standard deviation of $2.40 per hour and compute the p-value. c. With α = .05 as the level of significance, what is your conclusion? d. Repeat the preceding hypothesis test using the critical value approach.

Answer 1

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = $24.57  

    Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ $24.57

(b) The P-value of the test statistics is 0.1212.

(c) We conclude that the population mean hourly wage in the manufacturing industry equals the population mean hourly wage in the U.S industries using P-value approach.

(d) We conclude that the population mean hourly wage in the manufacturing industry equals the population mean hourly wage in the U.S industries using critical value approach.

Step-by-step explanation:

We are given that a sample of employees from the manufacturing industry to see if the mean hourly wage differs from the reported mean of $24.57 for the U.S industries.

Suppose a sample of 30 employees from the manufacturing industry showed a sample mean of $23.89 per hour. Assume a population standard deviation of $2.40 per hour.

Let $\mu$ = population mean hourly wage in the manufacturing industry.

(a) Null Hypothesis, $H_0$ : $\mu$ = $24.57     {means that the population mean hourly wage in the manufacturing industry equals the population mean hourly wage in the U.S industries}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ $24.57     {means that the population mean hourly wage in the manufacturing industry differs from the population mean hourly wage in the U.S industries}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

                            T.S. =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\bar X$ = sample mean wage in the manufacturing industry = $23.89/hr

            σ = population standard deviation = $2.40/hr

            n = sample of employees from the manufacturing industry = 30

So, the test statistics  =  $\frac{23.89-24.57}{\frac{2.40}{\sqrt{30} } }$

                                      =  -1.55

The value of z test statistics is -1.55.

(b) Now, the P-value of the test statistics is given by;

                P-value = P(Z < -1.55) = 1 - P(Z $\leq$ 1.55)

                              = 1 - 0.9394 = 0.0606

For two-tailed test P-value is calculated as = 0.0606 $\times$ 2 = 0.1212

Since, the P-value of the test statistics is higher than the level of significance as 0.1212 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the population mean hourly wage in the manufacturing industry equals the population mean hourly wage in the U.S industries.

(d) Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the population mean hourly wage in the manufacturing industry equals the population mean hourly wage in the U.S industries.