Question

(1 point) 2.8070 An automobile manufacturer would like to know what proportion of its customers are not satisfied with the service provided by the local dealer. The customer relations department will survey a random sample of customers and compute a 99.5% confidence interval for the proportion who are not satisfied. (a) Past studies suggest that this proportion will be about 0.15. Find the sample size needed if the margin of the error of the confidence interval is to be about 0.025. (You will need a critical value accurate to at least 4 decimal places.) Sample size: (b) Using the sample size above, when the sample is actually contacted, 16% of the sample say they are not satisfied. What is the margin of the error of the confidence interval? MoE:

Answer 1

Solution :

(a)

we have p = 0.15, q = 1 - p = 1-0.15 = 0.85, Margin of error (ME) = 0.025 and z critical value for 99.5% confidence level is 2.807 using normal distribution table

we have to find the sample size

formula for sample size is given as

n = (p*q)*[z/ME]^2

setting the given values, we get

n = (0.15*0.85)[2.807/0.025]^2 = 1607. 366or 1607

So, the required sample size is 1607

(b)

Now, we have sample size n = 1607, p = 0.16, q = 1-0.16 = 0.84 and z= 2.807 for 99.5% confidence level

Now, we have to find the value of margin of error

using the formula

ME =

Z*sq((p*q) /n) =2.807*sq((0.16*0.84)/1607

=0.02567

= 0.026

so, the required margin of error is 0.026

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