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Seventy-three percent of a sample of 400 customers surveyed had a favourable impression of a new product (a) Estimate the val

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Answer)

A)

P = 0.73

N = sample size = 400

Value of the proportion = n*p = 400*0.73 = 292

B)

Standard error = √p*(1-p)/√n

= √{0.73*(1-0.73)}/√400

Standard error = 0.0221979728804

C)

First we need to check the conditions of normality

That is if n*p and n*(1-p) both are greater than 5 or not

N*p = 292 > 5

N*(1-p) = 108 > 5

As both the conditions are met, we can use standard normal z table to estimate the interval

From z table, critical value z for 95% confidence level is 1.96.

Margin of error = z*standard error = 1.96*0.0221979728804

MOE = 0.0435080268456

Confidence interval is given by,

(P-MOE, P+MOE)

(0.6864919731543, 0.7735080268456)

D)

We are 95% confident that true population proportion lies in the interval (0.686, 0.774)

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