5. Scenario: A survey is conducted to determine the average number of people living in a...

5. Scenario: A survey is conducted to determine the average number of people living in a household in Scranton. An interviewer asks a sample of 16 Scranton shoppers, selected randomly in the Viewmont Mall (no two living in the same household), how many people live in their household. The results are as follows: 1 1 1 2 2 2 2 2 2 3 3 3 4 4 5 6

Assume the distribution of the number in all households is approximately normal. Do each of the following:

Determine the mean and standard deviation for the sample data from the survey.

x̄ = ______________ s = __________________

Using the sample results in Part (a), find a 95% confidence interval for the mean number of members in all Scranton households.

We can be _____________________________________________________________________

(c ) Given that the census bureau finds the standard deviation of all households sizes in Scranton is 1.28, find the number of sample items required for a 99% confidence level for the mean of the population to be accurate within a margin of error of 0.1.

n = _______________

Solutions

Expert Solution

from above :xbar =2.688 ; s =1.448

95% Confidence interval for mean =1.916 ; 3.459

for 99 % CI value of z=	2.576
standard deviation σ=	1.2800
margin of error E =	0.1
required sample size n=(zσ/E)² =	1088.0

orchestra answered 1 year ago

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