Question

A restaurant manager asks all of the customers this month to take an online survey (and get a free appetizer their next visit). It turns out that 241 customers actually take the survey, and of these 209 were ''extremely satisfied'' with their visit. Set up an approximate 95% confidence interval for the proportion of all the restaurant's customers that say they were ''extremely satisfied'' with their visit..

Select only one of the boxes below.

A. The Normal curve cannot be used to make the requested confidence interval.

B. Making the requested confidence interval does not make any sense.

C. It is appropriate to compute a confidence interval for this problem using the Normal curve.

If it is appropriate to compute a confidence interval for this problem using the Normal curve, then enter the confidence interval below. Otherwise, enter [0,0] for the confidence interval.

Answer 1

(a)

Conditions for Normal Distribution approximation to Binomial Distribution:

n = Sample Size = 241

p = Sample Proportion = 209/241 = 0.8672

q = 1 - p = 0.1328

(i) np = 241 X 0.8672 = 209 > 10

and

(2) n (1 - p) = 241 X (1 - 0.8672) = 32 > 10

Both conditions are satisfied.

So,

Normal Distribution approximation can be applied to Binomial Distribution

So,

Correct option:

C. It is appropriate to compute a confidence interval for this problem using the Normal curve.

(b)

n = Sample Size = 241

p = Sample Proportion = 209/241 = 0.8672

q = 1 - p = 0.1328

SE =

= 0.05

From Table, critical values of Z = 1.96

Confidence Interval:
0.8672 (1.96 X 0.0219)

= 0.8672 0.0428

= [ 0.8244 ,0.9100]

So,

Confidence Interval:

[0.8244 ,0.9100]