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# 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. 175 likes 877 views ## 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.

Hermann
4.6
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.
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