the data accompanying this exercise show hourly wages (wage in $) for 50 employees. an economist wants to test if the average hourly wage is less than $28. assume that the population standard deviation is $7.



Answer :

The null hypothesis is not rejected, there is not enough evidence to claim that the population mean is less than 29.

Here, we are the data as shown in the figure below.

The hypothesis to be tested here will be-

H₀ ⇒ μ = 29

H₁ ⇒ μ < 29

The significance level α is given to be 0.01

From the z tables we can calculate the critical value as z as -1.282

The test statistic is calculated as-

z = (X - μ)[tex]\sqrt{n}[/tex]/ σ

where X is the sample mean

n =  the number of observations

and σ = sample standard deviation

Substituting the values into the formula we get-

z = (28.18 - 29)[tex]\sqrt{50}[/tex]/ 8

z = -0.725

Now, we can observe that -0.725 > -1.28 ⇒ test statistic is greater than the critical value

Thus, we can conclude that the null hypothesis is not rejected.

Hence, there is not enough evidence to claim that the population mean is less than 29.

Learn more about hypothesis testing here-

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