The mean hourly wage for employees in goods-producing industries was $24.57, April 12, 2012 (Bureau of Labor Statistics). Suppose a sample of 30 employees from the manufacturing industry is taken, in April 2012, to see if the mean hourly wage differs from the reported mean of $24.57 for the goods-producing 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 goods-producing industries.
b. Suppose a sample of 30 employees from the manufacturing industry showed a sam- ple mean of $23.89 per hour. Assume a population standard deviation of $2.40 per hour and compute the p-value.
c. With a = .05 as the level of significance, what is your conclusion?
d. Repeat the preceding hypothesis test using the critical value approach.

Step-by-step explanation:

a. The null hypothesis is generally an exact value and the alternative hypothesis is the one we are trying to show. We are trying to show that the population mean is different from $24.57, so the hypotheses are as follows:

Null: The population mean hourly wage in the manufacturing industry is the same as the population mean hourly wage in the goods-producing industries

Alternative: The population mean hourly wage in the manufacturing industry differs from the population mean hourly wage in the goods-producing industries

b.

Because we know the population standard deviation, we can use a z test

Plugging in the values:

Expected population mean: 24.57

Sample average: 23.89

Sample size: 30

Population standard deviation: 2.4, we get 0.1207 as our p-value.

c. Using a = 0.05, our p-value is greater than that, so we can not conclude the alternative hypothesis (The population mean hourly wage in the manufacturing industry differs from the population mean hourly wage in the goods-producing industries)

d. Using a critical value calculator, we can find that for a two-tailed approach, the critical value would be 1.96. This would mean that the z score would have to be greater than 1.96 or less than -1.96 for the results to be significant. The population mean we would plug in here is 24.57, while the raw score would be 23.89 and the standard deviation would be 2.4. The z score we get is -0.283, which is not in the values specified (>1.96 or <-1.95) so we cannot conclude the alternative hypothesis

17.08

correct me if i am wrong.

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497