Question

Economists often track employment trends by measuring the proportion of people who are "underemployed," meaning they are either unemployed or would like to work full time but are only working part-time. In the summer of 2013, 17. 6% of Americans were "underemployed. " The mayor of Thicksburg wants to show the voters that the situation is not as bad in his town as it is in the rest of the country. His staff takes a simple random sample of 300 Thicksburg residents and finds that 45 of them are underemployed.


Does the data give convincing evidence at the a = 0. 05 level that the proportion of underemployed residents in Thicksburg is lower than elsewhere in the country?

Answer 1

The answer to whether the data is convincing is; Yes, it is true that the data is convincing

How to create hypothesis?

We are given;

Population Proportion; p = 17.6% = 0.176

Sample size; n = 300

Sample proportion; p^ = 45/300 = 0.15 = 15%

Formula for standard deviation here is;

σ = √(pq/n)

σ = √(p(1 - p)/n)

σ =  √(0.15 * 0.85/300)

σ = 0.0206

Formula for z-score is;

z = (p^ - p)/σ

z = (0.176 - 0.15)/0.0206

z = 1

From online p-value from z-score at significance level of 0.05, we have;

p = 0.1587

The p-value is greater than the significance value and so we fail to reject the null hypothesis and conclude that  the proportion of underemployed residents in Thicksburg is lower than elsewhere in the country.

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