Question

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.

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Answer to a math question 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.

Expert avatar
Hermann
4.6
65 Answers
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.

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