A drug company has developed a new vaccine for preventing the flu. The company claims that fewer than 5% of adults who use its vaccine will get the flu. To test the claim, researchers give the vaccine to a random sample of 1000 adults. Of these, 43 get the flu. Do these data provide convincing evidence to support the company’s claim? Perform an appropriate test to support your answer. Which kind of mistake (Type 1 or Type 2) could you have made in part a? Explain. From the company’s point of view, would Type 1 or Type 2 error be more serious? Why?

Step-by-step explanation:

We would set up the hypothesis test.

For the null hypothesis,

p ≤ 0.05

For the alternative hypothesis,

p > 0.05

This is aright tailed test.

Considering the population proportion, probability of success, p = 0.05

q = probability of failure = 1 - p

q = 1 - 0.05 = 0.95

Considering the sample,

Sample proportion, P = x/n

Where

x = number of success = 43

n = number of samples = 1000

p = 43/1000 = 0.043

We would determine the test statistic which is the z score

z = (P - p)/√pq/n

z = (0.043 - 0.05)/√(0.05 × 0.95)/400 = - 0.64

Since the test is a right tailed test, we would find the probability value for the area above the z score. It becomes

p = 1 - 0.26 = 0.74

Let us assume a significance level of 0.05

Since alpha, 0.05 < than the p value, 0.74, then we would fail to reject the null hypothesis. A type I error occurs when a true null hypothesis is rejected. A type 2 error occurs when a false null hypothesis is accepted.

A type 2 error could have been made in part A.

A type 2 error would be more serious because people will believe that the vaccine is very effective in preventing the flu when it is not. This can lead to more adults getting the flu.

A type 1 error.

Because we could have rejected a true null hypothesis and thus support the company's claim.

Step-by-step explanation:

Sample size (n) = 1000

Take the following hypothesis:

Null hypothesis: p = 0.05

Alternative hypothesis: p < 0.05

Po = 0.05

Sample proportion(P) = (number of success/sample size) = 43/1000 = 0.043

Using the t-test statistic :

Obtain the z-score using the formula:

z = (P - Po) / √[Po(1 - Po) / n]

z = (0.043 - 0.05) / √[0.05(1 - 0.05) / 1000]

z = -0.007 / √(0.0475 / 1000)

z = -0.007 / √0.0000475

z = −1.015666

Calculating the P-value of the z-score obtained using the p-value calculator :

P = P(z < −1.015666) = 0.1548942

Since the p-value is greater than the significance level, we cannot reject the Null hypothesis.

P > 0.05, Therefore we fail to reject the null hypothesis as there is not enough evidence to support the company's claim.