The mean hourly wage for employees in goods-producing industries is currently $2

Question

The mean hourly wage for employees in goods-producing industries is currently $24.57 (Bureau of Labor Statistics website, April, 12, 2012). 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 goods-producing industries. a. State the null 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. 1. 2. 3. Choose correct answer from above choice State the 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. 1. 2. 3. Choose correct answer from above choice 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. Round your answer to four decimal places. 24.57 c. With = .05 as the level of significance, what is your conclusion? The input in the box below will not be graded, but may be reviewed and considered by your instructor. blank d. Repeat the preceding hypothesis test using the critical value approach. Round your answer to two decimal places. Enter negative values as negative numbers. z = 1.996 ; H0

Explanation / Answer

Given that,
population mean(u)=24.57
standard deviation, =2.4
sample mean, x =23.89
number (n)=30
null, Ho: =24.57
alternate, H1: !=24.57
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 23.89-24.57/(2.4/sqrt(30)
zo = -1.55
| zo | = 1.55
critical value
the value of |z | at los 5% is 1.96
we got |zo| =1.55 & | z | = 1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -1.55 ) = 0.12
hence value of p0.05 < 0.12, here we do not reject Ho
ANSWERS
---------------
null, Ho: =24.57
alternate, H1: !=24.57
test statistic: -1.55
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.12

no evidence to support the mean hourly wage differs from the reported mean of $24.57

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