Question

A fast food company CEO claims that 19% of university students regularly eat at one of their restaurants. A survey of 145 students showed that only 21 students regularly eat at one of their restaurants. Assuming the CEO's claim is correct, determine (to 4 decimal places):

1. the standard error for the sampling distribution of the proportion.
2. the probability that the sample proportion is no more than that found in the survey.

Answer 1

Given in the most of the statistical books as well as the practical real world, if you really have a large enough sample size, you would just make inferences about the population proportions. And also you can then use the normal distribution for the sampling distribution of P-cap, P^, if you have given large sample size.

Use the formulae given below for the basic guideline on "How large is considered 'large enough'?"

1. np is greater than equal to 5.

2. n(1-p) is greater than equal to 5.

Check that your sample size is large enough :

In this case, you've a sample size of 145 students and a proportion of 19%, now as we follow the above formulated steps, we get,

Step 1) n*p = 145*0.19 = 27.55

Step 2) n(1-p) = 145 (1-0.19) = 145 (0.81) = 117.45

Both are greater (or above) than 5 and hence, we can now use the normal distribution.

Ans 1) The Standard Error for the Sampling Distribution of the proportion :

√ [p(1-p)/n] = √ [0.19 (1-0.19)/145]

which further implies √ [0.19 (0.81)/145]

= √ (0.1539/145) = √0.0010 =0.0316

Ans 2) Determination of the probability that the sample proportion is no more than that found in the survey :

z = (P1-p)/Standard Error

P(Z greater than equal to) (0.14-0.19)/0.0316 = (-)1.58

At a probability of 0.1145, it is likely that the proportion of University students who regularly eat at one of the restaurants of a Fast Food Company would have been 14% without a campaign.

Now, I found a value of (-)0.4429 in the z-table. That indicates the area between the mean and z. (Ignore the negative sign)