# 1. The manager of a large factory believes that the average hourly wage of the employees is below \$9.78 per hour. A sample of 18 employees has a mean hourly wage of \$9.60. The standard deviation of all salaries is \$1.42. Assume the variable is normally distributed. At  = 0.10, is there enough evidence to support the manager’s claim?

## Z =(score-mean)/SD = (9.60-9.78)/1.42 = ?

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. Compare with .10.

## To determine if there is enough evidence to support the manager's claim that the average hourly wage of the employees is below \$9.78 per hour, we can conduct a hypothesis test.

Step 1: State the null and alternative hypotheses:
- Null hypothesis (H0): The average hourly wage of the employees is \$9.78 or higher.
- Alternative hypothesis (Ha): The average hourly wage of the employees is below \$9.78.

Step 2: Determine the level of significance, α:
- The given level of significance is α = 0.10.

Step 3: Select the appropriate test statistic:
- Since we know the population standard deviation (σ), we can use the z-test.

Step 4: Formulate the decision rule:
- We will reject the null hypothesis if the z-statistic falls in the critical region, which is determined by the given α value.
- For a one-tailed test at α = 0.10, the critical z-value is -1.28 (from the standard normal distribution table).

Step 5: Calculate the test statistic:
- We can use the formula for calculating the z-statistic:
z = (x̄ - μ) / (σ / √n)
where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

In this case, x̄ = \$9.60, μ = \$9.78, σ = \$1.42, and n = 18.
Plugging in these values, we get:
z = (\$9.60 - \$9.78) / (\$1.42 / √18) = -0.18 / 0.335
z ≈ -0.54

Step 6: Make a decision:
- Since the calculated z-statistic (-0.54) is not less than the critical z-value (-1.28) for a one-tailed test at α = 0.10, we fail to reject the null hypothesis.

Step 7: State the conclusion:
- There is not enough evidence to support the manager's claim that the average hourly wage of the employees is below \$9.78 per hour at α = 0.10.

## To determine whether there is enough evidence to support the manager's claim that the average hourly wage is below \$9.78, we need to conduct a hypothesis test.

First, let's state the null and alternative hypotheses:

- Null hypothesis (H0): The average hourly wage is equal to or above \$9.78 per hour.
- Alternative hypothesis (Ha): The average hourly wage is below \$9.78 per hour.

Next, we need to calculate the test statistic, which is the t-score in this case. The formula for the t-score is:

t = (sample mean - population mean) / (sample standard deviation / √n)

Given the information in the question, we have:
- Sample mean (x̄) = \$9.60
- Population mean (µ) = \$9.78
- Sample standard deviation (s) = \$1.42
- Sample size (n) = 18

Plugging these values into the formula, we get:

t = (\$9.60 - \$9.78) / (\$1.42 / √18)

Calculating this expression will give us the t-score.

Once we have the t-score, we can compare it to the critical value. The critical value corresponds to the chosen significance level (α) and the degrees of freedom (n - 1). In this case, α = 0.10 and the degrees of freedom = 18 - 1 = 17.

Using a t-table or a statistical calculator, find the critical value for a two-tailed test with α = 0.10 and 17 degrees of freedom.

If the calculated t-score is greater than the critical value (in absolute value), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

By following these steps, you can determine whether there is enough evidence to support the manager's claim.