A random sample of 384 people in a mid-sized city (city one) revealed 112 individuals who worked at more than one job. A second random sample of 432 workers from another mid-sized city (city two) found 91 people who work at more than one job. Find a 99% confidence interval for the difference between the proportions of workers in the two cities who work at more than one job.
(0.003, 0.159)
(0.021, 0.141)
(-0.159, 0.004)
(0.031, 0.131)
Sample sizes aren't large enough to justify using z-procedures
Let's look at your data:
n1 = 384
n2 = 432
p1 = 112/384
p2 = 91/432
Formula:
CI99 = (p1 - p2) ± 2.58 √(p1(1-p1)/n1 + p2(1-p2)/n2)
Substitute the values into the formula and calculate. (Convert all fractions to decimals.)
You should be able to select your answer once you have determined the interval.
The answer is A, Math guru know what he be doin!
Why did the city one worker go to the city two worker's party? Because they heard there would be a 'proportion' of workers who work at more than one job! But don't worry, I'll help you find that confidence interval.
To find the confidence interval for the difference between the proportions, we can use the formula:
CI = (p1 - p2) +/- z*sqrt(p1(1-p1)/n1 + p2(1-p2)/n2)
where p1 and p2 are the proportions of workers in city one and city two, n1 and n2 are the sample sizes, and z is the critical value for a 99% confidence level.
Let's calculate it step by step, shall we?
First, let's calculate the proportions p1 and p2:
p1 = 112/384 = 0.2917
p2 = 91/432 = 0.2106
Next, we'll find the critical value z for a 99% confidence level. Since the sample sizes are large enough, we can use the standard normal distribution. The z value for a 99% confidence level is approximately 2.576.
Now, let's substitute the values into the formula:
CI = (0.2917 - 0.2106) +/- 2.576*sqrt(0.2917(1-0.2917)/384 + 0.2106(1-0.2106)/432)
After doing the calculations, we get:
CI = (0.0811) +/- 2.576*sqrt(0.002872 + 0.003279)
Simplifying this further:
CI = (0.0811) +/- 2.576*sqrt(0.006151)
CI = (0.0811) +/- 2.576*0.078429
CI = (0.0811) +/- 0.202001
Finally, the confidence interval is approximately:
CI = (0.003, 0.159)
So the correct answer is (0.003, 0.159), like a party with a confidence level and a difference in proportions!
To find a 99% confidence interval for the difference between the proportions of workers in the two cities who work at more than one job, we can use the formula:
CI = (p̂1 - p̂2) ± z * √(p̂1 * (1 - p̂1) / n1 + p̂2 * (1 - p̂2) / n2)
where:
p̂1 = proportion of workers in city one who work at more than one job
p̂2 = proportion of workers in city two who work at more than one job
n1 = sample size of city one
n2 = sample size of city two
z = z-score for a 99% confidence level
First, let's calculate the proportions:
p̂1 = 112/384 = 0.2917
p̂2 = 91/432 = 0.2106
Next, let's calculate the standard error:
SE = √(p̂1 * (1 - p̂1) / n1 + p̂2 * (1 - p̂2) / n2)
= √(0.2917 * (1 - 0.2917) / 384 + 0.2106 * (1 - 0.2106) / 432)
= 0.0269
Now, let's find the z-score for a 99% confidence level (two-tailed):
z = 2.576
Finally, let's calculate the confidence interval:
CI = (p̂1 - p̂2) ± z * SE
= (0.2917 - 0.2106) ± 2.576 * 0.0269
= 0.0811 ± 0.0692
= (0.0119, 0.1503)
Therefore, the 99% confidence interval for the difference between the proportions of workers in the two cities who work at more than one job is (0.0119, 0.1503).
The correct answer is not provided in the options.
To find the confidence interval for the difference between the proportions of workers in the two cities who work at more than one job, we can use the formula:
Confidence Interval = (p1 - p2) ± z * sqrt((p1(1-p1)/n1) + (p2(1-p2)/n2))
where:
p1 = proportion of workers in city one who work at more than one job
p2 = proportion of workers in city two who work at more than one job
n1 = sample size of city one
n2 = sample size of city two
z = z-score based on the confidence level (in this case, 99% confidence level)
First, calculate the proportions:
p1 = 112/384 ≈ 0.2917
p2 = 91/432 ≈ 0.2106
Next, calculate the standard error:
SE = sqrt((p1(1-p1)/n1) + (p2(1-p2)/n2))
= sqrt((0.2917(1-0.2917)/384) + (0.2106(1-0.2106)/432))
≈ 0.0275
Since the sample sizes are large enough, we can use a z-test instead of a t-test. To find the z-score for a 99% confidence level, we need to find the z-value that corresponds to a 0.995 probability (0.5 + 0.99/2 = 0.995).
Using a standard normal distribution table or a calculator, we find that the z-value is approximately 2.576.
Finally, substitute the values into the confidence interval formula:
Confidence Interval = (p1 - p2) ± z * SE
= (0.2917 - 0.2106) ± 2.576 * 0.0275
= 0.0811 ± 0.0709
Calculating the bounds of the interval:
Lower Bound = 0.0811 - 0.0709 ≈ 0.0102
Upper Bound = 0.0811 + 0.0709 ≈ 0.1518
Therefore, the 99% confidence interval for the difference between the proportions of workers in the two cities who work at more than one job is approximately (0.0102, 0.1518).
The correct answer choice is (0.021, 0.141).