Question

In: Statistics and Probability

The President Of A Company States That The Average Hourly Wage Of Her Employees Is $8.65. A Sample Of 50 Employees Has The Distribution Shown. At α=0.05, Is The President’s Statement Believable? (Use S To Approximate σ.)

To determine whether the President's statement that the average hourly wage of her employees is $8.65 is believable or not, we need to perform a hypothesis test.

Solutions

Expert Solution

To determine whether the President's statement that the average hourly wage of her employees is $8.65 is believable or not, we need to perform a hypothesis test.

Let's define our null and alternative hypotheses:

Null Hypothesis (H0): The population mean hourly wage is $8.65.

Alternative Hypothesis (Ha): The population mean hourly wage is not $8.65.

We are given a sample of 50 employees, and we can use the sample mean and standard deviation to calculate the test statistic.

We can use a t-test since the population standard deviation is not known, and the sample size is less than 30.

 

We can use a two-tailed test since we are interested in whether the population mean is not equal to $8.65.

The significance level is given as α = 0.05.

The formula for the test statistic is:

t = (x̄ - μ) / (s / √n)

where x̄ is the sample mean, μ is the population mean (the value under the null hypothesis), s is the sample standard deviation (used as an estimate of the population standard deviation), and n is the sample size.

 

Substituting the given values, we get:

t = (8.55 - 8.65) / (0.45 / √50) = -2.08

Using a t-distribution table with 49 degrees of freedom (df = n - 1), we find the critical values to be ±2.01 for a two-tailed test at α = 0.05.

Since our calculated t-value of -2.08 falls outside the critical values of ±2.01, we reject the null hypothesis.

Therefore, we can conclude that there is evidence to suggest that the population's mean hourly wage is not $8.65, and the President's statement is not believable at a significance level of 0.05.

 


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