Question

Mateus’s bank issued an advertisement saying that 90\%90%90, percent of its customers are satisfied with the bank’s services. Since he himself wasn't very satisfied, he suspected the ad is false. He surveyed a random sample of 808080 of the bank’s customers, and found that only 80\%80%80, percent were satisfied. Let's test the hypothesis that the actual percentage of satisfied customers is 90\%90%90, percent versus the alternative that the actual percentage is lower than that. The table below sums up the results of 100010001000 simulations, each simulating a sample of 808080 customers, assuming there are 90\%90%90, percent satisfied customers. According to the simulations, what is the probability of getting a sample with 80\%80%80, percent satisfied customers or less?

Answer 1

Answer:

The probability of getting a sample with 80% satisfied customers or less is 0.0125.

Step-by-step explanation:

We are given that the results of 1000 simulations, each simulating a sample of 80 customers, assuming there are 90 percent satisfied customers.

Let $\hat p$ = sample proportion of satisfied customers

The z-score probability distribution for the sample proportion is given by;

                                Z  =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, p = population proportion of satisfied customers = 90%

            n = sample of customers = 80

Now, the probability of getting a sample with 80% satisfied customers or less is given by = P( $\hat p \leq$ 80%)

  P( $\hat p \leq$ 80%) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ $\leq$ $\frac{0.80-0.90}{\sqrt{\frac{0.80(1-0.80)}{80} } }$ ) = P(Z $\leq$ -2.24) = 1 - P(Z < 2.24)

                                                                  = 1 - 0.9875 = 0.0125

The above probability is calculated by looking at the value of x = 2.24 in the z table which has an area of 0.9875.