In the past 10 years, the number of U.S. adults living without a spouse or partn
ID: 3048654 • Letter: I
Question
In the past 10 years, the number of U.S. adults living without a spouse or partner has climbed. We would like to test if the population proportion of unemployed working age Americans who live without a spouse/partner is lower than the population proportion of employed working age Americans who live without a spouse/partner, using a 10% significance level.
The following data was obtained from random samples.
“Living with spouse/partner?”
Group
With
Without
Total
1 = Unemployed working age Americans
403
247
650
2 = Employed working age Americans
144
96
240
Recall the Steps for Testing Hypotheses:
1) State the appropriate hypotheses (the significance level has been selected to be 10%).
2) Check the necessary assumption(s) (you may assume that you have two independent random samples).
3) Perform the appropriate test (include all supporting computations for the test statistic and p-value).
4) Give your statistical decision and provide a conclusion in context.
Organize your answer well and use labels for your Steps 1, 2, 3, and 4.
“Living with spouse/partner?”
Group
With
Without
Total
1 = Unemployed working age Americans
403
247
650
2 = Employed working age Americans
144
96
240
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: PUnemployed> PEmployed
Alternative hypothesis: PUnemployed < PEmployed
Note that these hypotheses constitute a one-tailed test.
Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a two-proportion z-test.
Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).
p = (p1 * n1 + p2 * n2) / (n1 + n2)
p = 0.3854
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.03676
z = (p1 - p2) / SE
z = - 0.54
where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.
Since we have a one-tailed test, the P-value is the probability that the z-score is less than - 0.54.
We use the Normal Distribution Calculator to find P(z < - 0.54) = 0.2946
Interpret results. Since the P-value (0.2946) is greater than the significance level (0.10), we cannot reject the null hypothesis.
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