Question

In: Statistics and Probability

(2 point) Refer to the following scenario. An epidemiologist is worried about the prevalence of the...

(2 point) Refer to the following scenario.

An epidemiologist is worried about the prevalence of the flu in East Vancouver and the potential shortage of vaccines for the area. She will need to provide a recommendation for how to allocate the vaccines appropriately across the city. She takes a simple random sample of 333 people living in East Vancouver and finds that 36 have recently had the flu.

For each of the following statements, specify whether the statement is a correct interpretation of the 95% confidence interval for the true proportion of East Vancouver residents who have recently had the flu.

A. 10.81% (36/333) of East Vancouver residents have recently had the flu. true/ false

B. There is a 95% probability that the true proportion of East Vancouver residents who have recently had the flu equals 36/333. true/false

C. If another random sample of 333 East Vancouver residents is drawn, there is a 95% probability that the sample proportion of East Vancouver residents who have recently had the flu equals 36/333. true/false

D. If many random samples of 333 East Vancouver residents are drawn, 95% of the resulting confidence intervals will contain the value of the true proportion of East Vancouver residents who have recently had the flu. true/false

E. If many random samples of 333 East Vancouver residents are drawn, 95% of the resulting confidence intervals will contain the value 36/333. true/false

Part 2

An epidemiologist is worried about the prevalence of the flu in East Vancouver and the potential shortage of vaccines for the area. She will need to provide a recommendation for how to allocate the vaccines appropriately across the city. She takes a simple random sample of 340 people living in East Vancouver and finds that 33 have recently had the flu.

Suppose that the epidemiologist wants to re-estimate the population proportion and wishes for her 95% confidence interval to have a margin of error no larger than 0.04. How large a sample should she take to achieve this? Please carry answers to at least six decimal places in intermediate steps.

Sample Size =

Solutions

Expert Solution

Sol:

95% of the confidence intervals contain true population proportion who had flue

confidence interval is given for population proportion not for sample proportion

p^=sample proprtion who had flu=x/n=36/333= 0.1081081= 0.1081081*100=10.81%

A. 10.81% (36/333) of East Vancouver residents have recently had the flu. true/ false

TRUE

B. There is a 95% probability that the true proportion of East Vancouver residents who have recently had the flu equals 36/333. true/false

FALSE

C. If another random sample of 333 East Vancouver residents is drawn, there is a 95% probability that the sample proportion of East Vancouver residents who have recently had the flu equals 36/333.true/false

FALSE

D. If many random samples of 333 East Vancouver residents are drawn, 95% of the resulting confidence intervals will contain the value of the true proportion of East Vancouver residents who have recently had the flu. true/false

FALSE

E. If many random samples of 333 East Vancouver residents are drawn, 95% of the resulting confidence intervals will contain the value 36/333. true/false

FALSE

An epidemiologist is worried about the prevalence of the flu in East Vancouver and the potential shortage of vaccines for the area. She will need to provide a recommendation for how to allocate the vaccines appropriately across the city. She takes a simple random sample of 340 people living in East Vancouver and finds that 33 have recently had the flu.

Suppose that the epidemiologist wants to re-estimate the population proportion and wishes for her 95% confidence interval to have a margin of error no larger than 0.04. How large a sample should she take to achieve this? Please carry answers to at least six decimal places in intermediate steps.

p^=x/n=33/340=0.09705882= 0.097059

E=0.04

z crit fro 95%==NORM.S.INV(0.975)= 1.959964

n=(Z/E)^2*p^*(1-p^)

=(1.959964/0.04)^2*0.097059*(1-0.097059)

n=210.4124

Sample Size =n=211


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