The Pew Research Center recently polled n= 1048 U.S drivers andfound that 69% enjoyed driving their automobiles.
(a) Construct a 95% confidence interval for the proportion of U.S.drivers who enjoy driving their automobiles.
(b) In 1991, Gallup Poll reported this percent to be 79%. Using thedata from this poll, test the claim that the percent of drivers whoenjoy driving their cars has declined since 1991. Report thelarge-sample z statistic and its P-value.
(a) 95% confidence interval: (0.65, 0.73)
(b) Large-sample z statistic: -2.45
P-value: 0.014
(a) To construct a 95% confidence interval for the proportion of U.S. drivers who enjoy driving their automobiles, we can use the formula:
CI = p̂ ± z * √(p̂ * (1 - p̂) / n)
where p̂ is the sample proportion (69% = 0.69), z is the critical value for a 95% confidence level (1.96 for a large sample), and n is the sample size (1048).
Substituting the values into the formula:
CI = 0.69 ± 1.96 * √(0.69 * (1 - 0.69) / 1048)
Calculating the values:
CI = 0.69 ± 1.96 * √(0.69 * 0.31 / 1048)
= 0.69 ± 1.96 * √(0.2139 / 1048)
= 0.69 ± 1.96 * 0.0142
= 0.69 ± 0.0279
Therefore, the 95% confidence interval for the proportion of U.S. drivers who enjoy driving their automobiles is (0.6621, 0.7179).
(b) To test the claim that the percent of drivers who enjoy driving their cars has declined since 1991, we can use the large-sample z-statistic and its P-value.
The large-sample z-statistic can be calculated using the formula:
z = (p̂ - p) / √(p * (1 - p) / n)
where p̂ is the sample proportion (0.69), p is the population proportion in 1991 (0.79), and n is the sample size (1048).
Substituting the values into the formula:
z = (0.69 - 0.79) / √(0.79 * (1 - 0.79) / 1048)
= (-0.10) / √(0.79 * 0.21 / 1048)
= (-0.10) / √(0.1659 / 1048)
= (-0.10) / √0.00015841
= (-0.10) / 0.0126
≈ -7.94
The large-sample z-statistic is approximately -7.94.
To find the P-value for this test, we can use a standard normal distribution table or a calculator. The P-value will be the probability of observing a z-value more extreme than -7.94 in the standard normal distribution.
The P-value for z = -7.94 is extremely small (close to 0). Therefore, we can reject the null hypothesis and conclude that the percent of drivers who enjoy driving their cars has declined since 1991.
To construct a confidence interval for the proportion of U.S. drivers who enjoy driving their automobiles, we can use the following formula:
Confidence interval = p̂ ± (z * √(p̂(1-p̂)/n))
Where:
- p̂ is the proportion of drivers who enjoy driving their automobiles (69% or 0.69 in this case),
- z is the z-score corresponding to the desired confidence level (95% in this case), and
- n is the sample size (1048 in this case).
(a) To find the z-score, we need to determine the critical value for a 95% confidence level. This can be done using a standard normal distribution table or software. The z-score for a 95% confidence level is approximately 1.96.
Now we can calculate the confidence interval:
Confidence interval = 0.69 ± (1.96 * √((0.69 * (1-0.69))/1048))
Confidence interval = 0.69 ± (1.96 * √((0.69 * 0.31)/1048))
Calculating this out, we find that the confidence interval is approximately (0.664, 0.716).
Therefore, we can say with 95% confidence that the true proportion of U.S. drivers who enjoy driving their automobiles is between 0.664 and 0.716.
(b) To test the claim that the percent of drivers who enjoy driving their cars has declined since 1991, we can conduct a hypothesis test. The null hypothesis (H0) is that the percent has not declined (p = 0.79), and the alternative hypothesis (Ha) is that the percent has declined (p < 0.79).
To perform this test, we can calculate the z statistic using the formula:
z = (p̂ - p) / √(p * (1 - p) / n)
Where:
- p̂ is the proportion in the sample (0.69),
- p is the proportion from the Gallup Poll in 1991 (0.79),
- n is the sample size (1048).
Calculating this out, we find:
z = (0.69 - 0.79) / √(0.79 * (1 - 0.79) / 1048)
z = -4.818
The large-sample z statistic is -4.818.
To determine the p-value, we can use a standard normal distribution table or software. The p-value is the probability of obtaining a z statistic as extreme as -4.818 or more extreme if the null hypothesis is true.
The p-value corresponding to a z statistic of -4.818 is less than 0.0001.
Therefore, we can conclude that the results from the Pew Research Center poll provide strong evidence to reject the null hypothesis. The proportion of drivers who enjoy driving their cars has declined significantly since 1991, as evidenced by the large negative z statistic and very small p-value.