A poll of 1,275 adults in the United States asked how much confidence they had in banks. A total of 166 adults said they had a great deal of confidence. An economist claims that less than 15% of U.S. adults have a great deal of confidence in banks. Can you conclude that the economist's claim is
True?
Use the $\alpha = 0.01$ level of significance.
A) No conclusion is possible.
B) Yes
C) No

Question

A poll of 1,275 adults in the United States asked how much confidence they had in banks. A total of 166 adults said they had a great deal of confidence. An economist claims that less than 15% of U.S. adults have a great deal of confidence in banks. Can you conclude that the economist's claim is
True? 
Use the $\alpha = 0.01$ level of significance.
A) No conclusion is possible.
B) Yes
C) No

Quizplus · Accepted Answer

We are given a sample size n = 1275, and from this sample 166 people said they have a great deal of confidence in banks. We can set up the null and alternative hypotheses as:
H0: p >= 0.15
Ha: p < 0.15
where p is the true proportion of adults in the US who have a great deal of confidence in banks. Since we are testing a claim about a proportion, we can use the normal approximation to the binomial distribution. The test statistic is:
z = (p̂ - p0) / sqrt(p0(1-p0)/n)
where p̂ is the sample proportion, p0 is the hypothesized proportion under the null hypothesis, and n is the sample size.

Under the null hypothesis, we have:
p̂ = 166 / 1275 = 0.13
z = (0.13 - 0.15) / sqrt(0.15(1-0.15)/1275) = -1.83

At the α = 0 .01 level of significance with a one-tailed test, the critical value is -2.33. Since the calculated test statistic (-1.83) is greater than the critical value (-2.33), we fail to reject the null hypothesis.

Therefore, we do not have sufficient evidence to conclude that the proportion of US adults who have a great deal of confidence in banks is less than 15%.

A Poll of 1,275 Adults in the United States Asked α=0.01\alpha = 0.01α=0.01

A Poll of 1,275 Adults in the United States Asked $\alpha = 0.01$