Concept explainers
Country Financial, a financial services company, uses surveys of adults age 18 and older to determine whether personal financial fitness is changing over time (USA Today, April 4, 2012). In February 2012, a sample of 1000 adults showed 410 indicating that their financial security was more than fair. In February 2010, a sample of 900 adults showed 315 indicating that their financial security was more than fair.
- a. State the hypotheses that can be used to test for a significant difference between the population proportions for the two years.
- b. What is the sample proportion indicating that their financial security was more than fair in 2012? In 2010?
- c. Conduct the hypothesis test and compute the p-value. At a .05 level of significance, what is your conclusion?
- d. What is the 95% confidence
interval estimate of the difference between the two population proportions?
a.
State the null and alternative hypotheses.
Answer to Problem 43SE
Null hypothesis:
Alternative hypothesis:
Explanation of Solution
Calculation:
It is given that in February 2012, 410 out of 1,000 adults indicated that their financial security was more than fair and in February 2010, 315 out of 900 adults indicated that their financial security was more than fair.
Here,
State the hypothesis:
Null hypothesis:
That is, there is no significant difference between the population proportions for the two years.
Alternative hypothesis:
That is, there is a significant difference between the population proportions for the two years.
b.
Find the sample proportion that indicates that their financial security was more than fair in 2012 and 2010.
Answer to Problem 43SE
The sample proportions that indicate that their financial security was more than fair in 2012 and 2010 are 0.41 and 0.35, respectively.
Explanation of Solution
Calculation:
Sample proportion for 2012:
The sample proportion for 2012
Thus, the sample proportion for 2012 is 0.41.
Sample proportion for 2010:
The sample proportion for 2010
Thus, the sample proportion for 2010 is 0.35.
c.
Find the p-value and provide a conclusion at
Answer to Problem 43SE
The p-value is 0.007.
There is sufficient evidence to conclude that there is a significant difference between the population proportions for the two years.
Explanation of Solution
Calculation:
The test statistic for the hypothesis test about
Here,
From Part (b),
Pooled estimator:
Substitute
Thus, the test statistic z-value is 2.69.
Software procedure:
Step-by-step procedure to obtain the probability value using Excel.
- Open an EXCEL sheet and select the cell A1.
- Enter the formula =NORM.S.DIST(2.69,TRUE) in the cell A1.
- Press Enter.
The output obtained using EXCEL software is given below:
From the output, the p-value for the left tail is 0.9964.
For the upper-tail test, the p-value is
From the output, the result is calculated as follows:
Thus, the p-value is 0.007.
Rejection rule:
If the
If the
Conclusion:
Here, the p-value is less than the level of significance.
That is,
From the rejection rule, the null hypothesis is rejected.
Therefore, there is sufficient evidence to conclude that there is a significant difference between the population proportions for the two years.
c.
Obtain a 95% confidence interval for the difference between the proportions of the two populations.
Answer to Problem 43SE
The 95% confidence interval is
Explanation of Solution
Calculation:
The formula for the confidence interval estimate of the difference between two populations is as follows:
Here,
Critical value:
For the 95% confidence level:
Hence, cumulative area to the left is calculated as follows:
From Table II of the standard normal distribution in Appendix B, the critical value is 1.96.
Confidence interval:
The 95% confidence interval for the difference between two population proportions is obtained as shown below:
Thus, the 95% confidence interval is
Interpretation:
There is 95% confidence that the population proportion of adults who say that their financial security is more than fair between 0.017 and 0.103.
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