The CEO of a large electric utility claims that 80 percent of his 1,000,000 customers are very satisfied with the service they receive. To test this claim, the local newspaper surveyed 100 customers, using simple random sampling. They are interested in seeing if this number over or under estimates the level of satisfaction. Among the sampled customers, 73 percent say they are very satisfied. Based on these findings, can we reject the CEO’s hypothesis that 80% of the customers are very satisfied?

1.Conduct a hypothesis test and calculate a 95% confidence interval for the estimate.

2. Suppose the question is slightly modified. If the CEO claims that at least 80 percent of the company’s 1,000,000 customers are very satisfied. Again, 100 customers are surveyed using simple random sampling. The result: 73 percent are very satisfied. Based on these results, should we accept or reject the CEO’s hypothesis?

Step-by-step explanation:

1) We would set up the hypothesis test.

For the null hypothesis,

P = 0.8

For the alternative hypothesis,

P ≠ 0.8

Considering the population proportion, probability of success, p = 0.8

q = probability of failure = 1 - p

q = 1 - 0.8 = 0.2

Considering the sample,

Sample proportion, p = x/n

Where

x = number of success = 73

n = number of samples = 100

p = 73/100 = 0.73

We would determine the test statistic which is the z score

z = (p - P)/√pq/n

z = (0.73 - 0.8)/√(0.8 × 0.2)/100 = - 1.75

Recall, population proportion, P = 0.8

The difference between sample proportion and population proportion is 0.8 - 0.73 = 0.07

Since the curve is symmetrical and it is a two tailed test, the p for the left tail is 0.8 - 0.07 = 0.73

the p for the right tail is 0.8 + 0.07 = 0.87

These proportions are lower and higher than the null proportion. Thus, they are evidence in favour of the alternative hypothesis. We will look at the area in both tails. Since it is showing in one tail only, we would double the area

From the normal distribution table, the area below the test z score in the left tail 0.04

We would double this area to include the area in the right tail of z = 1.75 Thus

p = 0.04 × 2 = 0.08

At 95% confidence level,

α = 1 - 0.95 = 0.05

Since α, 0.05 < than the p value, 0.08, then we would fail to reject the null hypothesis. Therefore, based on these findings, we cannot reject the CEO’s hypothesis that 80% of the customers are very satisfied.

Confidence interval is written as

Sample proportion ± margin of error

Margin of error = z × √pq/n

Where

z represents the z score corresponding to the confidence level

p = sample proportion. It also means probability of success

q = probability of failure

q = 1 - p

p = x/n

Where

n represents the number of samples

x represents the number of success

From the information given,

n = 73

x = 100

p = 73/100 = 0.73

q = 1 - 0.73 = 0.27

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, confidence level of 95% is 1.96

Therefore, the 95% confidence interval is

0.73 ± 1.96√(0.73)(0.27)/100

= 0.73 ± 0.087

2) We would set up the hypothesis test.

For the null hypothesis,

P ≥ 0.8

For the alternative hypothesis,

P < 0.8

This is a left tailed test

From the previous calculation, z = - 1.75

The p value is the area to the left of z = - 1.75

P value = 0.04

α = 1 - 0.95 = 0.05

Since α, 0.05 > than the p value, 0.04, then we would reject the null hypothesis. Therefore,

Based on these results, we should reject the CEO’s hypothesis

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