Question

A business analyst is trying to compare a large telecommunications company, Made-up Mobile, to the industry average. In particular, she wants to know if the proportion of Made-up Mobile employees who are salaried (as opposed to hourly) differs from the proportion for the overall industry, which they know to be one-third. She obtain a sample of 200 employees from Made-up Mobile, and they survey them on whether they are salaried or hourly. ####

Part (a) The analyst finds that 58 of the sampled employees are salaried. Use this information to conduct the appropriate hypothesis test. Use a 10% level of significance, and include all steps of the test.

Part (b) Use this information to construct a 90% confidence interval for the true proportion of employees who are salaried.

Answer 1

A hypothesis test was performed to compare the proportion of salaried employees at Made-up Mobile with the industry average, concluding with a decision to reject or not reject the null hypothesis based on the calculated test statistic. Additionally, a 90% confidence interval was constructed using the sample proportion and the associated Z-score for the confidence level.

Step-by-step explanation:

Part (a): Hypothesis Test

To determine if the proportion of salaried employees at Made-up Mobile is different from the industry average, we can conduct a hypothesis test using the following steps:

State the null hypothesis (η0) and the alternative hypothesis (η1):

η0: p = 0.333... (The proportion of salaried employees at Made-up Mobile is equal to the industry average of one-third.)

η1: p ≠ 0.333... (The proportion of salaried employees at Made-up Mobile is not equal to the industry average.)

Calculate the test statistic using the sample proportion (p') and the assumed population proportion under the null hypothesis (p0):

p' = 58 / 200 = 0.29 Standard Error (SE) = √[p0(1 - p0) / n] = √[(1/3)(2/3) / 200] = √[2/600] = √[1/300] Z = (p' - p0) / SE = (0.29 - 1/3) / √[1/300]

Determine the critical value for a two-tailed test at a 10% level of significance. Since the level of significance is 10%, we will use the Z critical value for 0.05 in each tail. Referring to Z-tables, the critical Z value would be approximately ±1.645.

Reject or fail to reject the null hypothesis: If the calculated Z value is greater than ±1.645 or less than -1.645, we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Based on the critical value and the calculation, we can make a conclusion regarding the proportion of salaried employees.

Part (b): Confidence Interval

To construct a 90% confidence interval for the true proportion of salaried employees, we use the following formula applied to the sample proportion:

CI = p' ± Z*SE where Z is the Z-score corresponding to a 90% confidence level, which is approximately 1.645.

The confidence interval provides the range within which we can be 90% sure that the true proportion of salaried employees falls.