The average weekly earnings in the leisure and hospitality industry group for a re‐ cent year was $273. A random sample of 40 workers showed weekly average ear‐ nings of $285 with the population standard deviation equal to 58. At the 0.05 level of significance can it be concluded that the mean differs from $273? Find a 95% con‐ fidence interval for the weekly earnings and show that it supports the results of the hypothesis test.

To determine whether the mean weekly earnings differs from $273, we will conduct a one-sample z-test.

Let's denote the population mean as μ, the sample mean as x̄, the population standard deviation as σ, the sample size as n, and the level of significance as α.

Given:
x̄ = $285
σ = $58
n = 40
α = 0.05

Step 1: State the null hypothesis (H0) and the alternative hypothesis (Ha):
H0: The mean weekly earnings, μ, is equal to $273.
Ha: The mean weekly earnings, μ, differs from $273.

Step 2: Calculate the test statistic.
We will use the formula for the z-test statistic:
z = \frac{x̄ - μ}{\frac{σ}{\sqrt{n}}}

Substituting in the given values:
z = \frac{285 - 273}{\frac{58}{\sqrt{40}}}

Step 3: Determine the critical value.
Since we are conducting a two-tailed test at the 0.05 level of significance, we need to find the critical z-value for α/2 = 0.025. This value can be obtained from the standard normal distribution table or calculator.

The critical z-value for a 0.025 level of significance is approximately ±1.96.

Step 4: Make a decision.
If the test statistic z falls outside the critical region (greater than 1.96 or less than -1.96), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Step 5: Calculate the p-value.
The p-value is the probability of obtaining a test statistic as extreme or more extreme than the observed value, assuming the null hypothesis is true.

To calculate the p-value, we can use a standard normal distribution table or a calculator. The p-value is the area under the standard normal curve that corresponds to the absolute value of the test statistic.

Step 6: Determine the conclusion.
If the p-value is less than the level of significance (α), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Now let's perform the calculations:

z = \frac{285 - 273}{\frac{58}{\sqrt{40}}} = \frac{12}{9.176 } ≈ 1.31

The critical z-value at the 0.025 level of significance is ±1.96.

Since the test statistic z (1.31) falls within the range -1.96 to 1.96, we fail to reject the null hypothesis.

Now, let's proceed to find the 95% confidence interval for the weekly earnings.

The formula for the confidence interval is:

\text{Confidence Interval}=\mu_0\pm z\frac{σ}{\sqrt{n}}

Plugging in the given values:

\text{Confidence Interval}=273\pm1.96\frac{58}{\sqrt{40}}

Calculating this, we get:

\text{Confidence Interval}=273\pm17.974

Thus, the 95% confidence interval for the weekly earnings is [255.026 : 290.974]. The population mean ($273) is within the confidence interval. Therefore, we cannot reject the null hypothesis.

Answer: No, at the 0.05 level of significance, it cannot be concluded that the mean weekly earnings differ from $273.