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.

A one sample z-test can be performed to determine whether the mean hourly wage differs from the reported mean of $24.57 for the goods-producing industries.

The hypothesis is defined as:

H₀: The mean hourly wage is same as the reported mean of $24.57 for the goods-producing industries, i.e. μ = $24.57.

Hₐ: The mean hourly wage differs from the reported mean of $24.57 for the goods-producing industries, i.e. μ ≠ $24.57.

The information provided is:

[tex]\bar x=\$23.89\\n=30\\\sigma=\$2.40[/tex]

Compute the test statistic as follows:

[tex]z=\frac{\bar x-\mu}{\sigma/\sqrt{n}}=\frac{23.89-24.57}{2.40/\sqrt{30}}=-1.55[/tex]

The test statistic value is, z = -1.55.

Compute the p-value of the test as follows:

[tex]p-value=2\times P(Z<-1.55)\\=2\times [1-P(Z<1.55)]\\=2\times (1-0.9394)\\=0.1212[/tex]

*Use a z-table for the probability.

Thus, the p-value of the test is 0.1212.