Suppose that in studies of mobile phone use while driving conducted in 2015 and 2020, 3.8% of the drivers observed in 2015 and 3.0% of the drivers observed in 2020 were using a handheld mobile phone while driving. Suppose that these percentages were based on observations from independent random samples of 1,500 drivers in 2015 and 1,500 drivers in 2020
(a) Are the sample sizes large enough to use the large-sample confidence interval for a difference in population
proportions?
(b) Assume that it is reasonable to regard these samples as representative of drivers in 2015 and drivers in 2020.
Estimate the difference in the proportion of drivers in 2015 and the proportion of drivers in 2020 who use a mobile
phone while driving using a 95% confidence interval. (Use 2015 Drivers − 2020 Drivers. Enter your answer using
interval notation. Round your numerical values to four decimal places.)
(c) Is 0 included in the confidence interval? What does this suggest about the difference in proportions for 2015 and 2020?
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(a) To determine whether the sample sizes are large enough to use the large-sample confidence interval for a difference in population proportions, we need to check if the conditions for using this interval are satisfied.
The conditions for using the large-sample confidence interval for a difference in population proportions are:
1) Both samples are random and representative of their respective populations.
2) The number of successes and failures in each sample is at least 10.
In this case, the samples are said to be random and independent. The problem states that they are also representative of their respective populations. Therefore, the first condition is satisfied.
To check the second condition, we need to calculate the number of successes and failures in each sample.
In 2015:
Number of successes = 0.038 * 1500 = 57
Number of failures = 1500 - 57 = 1443
In 2020:
Number of successes = 0.03 * 1500 = 45
Number of failures = 1500 - 45 = 1455
Both samples have more than 10 successes and failures, so the second condition is satisfied.
Therefore, the sample sizes are large enough to use the large-sample confidence interval for a difference in population proportions.
(b) To estimate the difference in the proportion of drivers in 2015 and the proportion of drivers in 2020 who use a mobile phone while driving, we can use the formula:
Difference in proportions = (proportion in 2015) - (proportion in 2020)
Standard Error = sqrt((proportion in 2015)(1 - proportion in 2015)/n1 + (proportion in 2020)(1 - proportion in 2020)/n2)
where n1 and n2 are the sample sizes.
Using the given values:
Proportion in 2015 = 0.038
Proportion in 2020 = 0.03
n1 = 1500
n2 = 1500
Difference in proportions = 0.038 - 0.03 = 0.0080
Standard Error = sqrt((0.038)(1 - 0.038)/1500 + (0.03)(1 - 0.03)/1500) = 0.0087
To construct a 95% confidence interval, we use the formula:
Estimate ± (Z * Standard Error)
where Z is the z-score corresponding to a 95% confidence level. For a 95% confidence level, Z is approximately 1.96.
Estimate = Difference in proportions = 0.0080
Z = 1.96
Standard Error = 0.0087
95% confidence interval = 0.0080 ± (1.96 * 0.0087)
= 0.0080 ± 0.0171
= (-0.0091, 0.0251)
Therefore, the 95% confidence interval for the difference in proportions of drivers in 2015 and 2020 who use a mobile phone while driving is (-0.0091, 0.0251) in interval notation.
(c) To check if 0 is included in the confidence interval, we compare the lower and upper bounds of the interval to 0.
Lower bound = -0.0091
Upper bound = 0.0251
Since 0 is between the lower and upper bounds, 0 is included in the confidence interval.
This suggests that there is no significant difference in the proportions of drivers in 2015 and 2020 who use a mobile phone while driving. The difference could be due to sampling variability.
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