old people often have a hard time finding work. AARP claims that it takes a work

Question

old people often have a hard time finding work. AARP claims that it takes a worker aged 55 plus 28 weeks to find a job. A random sample of individuals aged 55 plus was selected and they were asked how long it took them to find a job. The data are given in the table below. (AARP Bulletin, April 2008). 12 points) JobSearchTime weeks) 21-14?51T16-17. 14116112 480127117132 24 Job Search Time (Weeks) 12 10 52 21 26 14 13 24 19 28 26 26 10 21 Job Search Time (Weeks) 44 36 22 39 17 17 10 19 16 22 5 22 Provide a point estimate of the population mean number of weeks it takes a worker aged 55 plus to find a job. a) b) At 95% confidence level, what is the margin of error? Explain what it means in the context of this problem. c) what is 95% confidence interval estimate of the mean? Interpret what it means in this context. Using the confidence interval in part (c), is there any evidence to suggest that the population mean amount of time it takes a worker aged 55 plus is different from 28 weeks? d) e) Interpret the confidence level in this context? 1) What sample size would be required to obtain a margin of assumption will you make for this data set? error of 2 weeks job search? What

Explanation / Answer

Variable   N   Mean StDev SE Mean      95% CI
C1        40 22.00 11.89     1.88    (18.20, 25.80)

a)

sample mean is point estimate for population mean

here

sample mean = 22

b)

margin of error = t * sd/sqrt(n)

n = 40 , sd= 11.89 , degree of freedom = n-1 = 39

t =

=T.INV.2T(0.05,39) = 2.0227

hence margin of error = 2.0227 * 11.89/sqrt(40) = 3.802623

The margin of error is supposed to measure the maximum amount by which the sample results are expected to differ from those of the actual population

c)

95 % confidence interval

(sample mean +- margin of error)

= (18.20, 25.80)

d)

since 28 is not present in confidence interval, there is evidence that mean is different than 28 weeks

e)

we are 95 % confident that actual number of weeks will lie in given confidence interval

f)

we assume that sd = 11.89 is population sd

n = z^2 * sd^2 /e^2

= 1.96^2 * 11.89^2/2^2

= 135.7737

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