Question

12 Among 500 randomly selected adults were surveyed, there are 25% of them said that they used the Internet for shopping at least a few times a year. Construct a 90% confidence interval estimate of the percentage of adults who use the Internet for shopping.

Select one: a. (20.0%, 30.0%) b. (26.0%, 34.0%) c. (21.8%, 28.2%) d. (24.7%, 31.3%) e. None of the other answers is true.

14 To study the effects of prenatal cocaine use on infants, on collects a random sample of 200 weights of babies born to cocaine users that gives a mean of 2750 g and a standard deviation of 650 g. Use a 2% significance level to test the claim that weights of babies born to cocaine users have a mean that is less than 2900 g. Which one of the following is true?

Select one:

a. The test statistic is -3.26 and the critical value is -2.05, hence we reject the null hypothesis.

b. The test statistic is -3.26 and the critical value is -2.33, hence we reject the null hypothesis.

c. The test statistic is -3.26 and the critical value is 2.33, hence we reject the null hypothesis.

d. The test statistic is -1.09 and the critical value is -2.05, hence we fail to reject the null hypothesis.

e. None of the other answers in necessary true.

Answer 1

12)

sample proportion, = 0.25
sample size, n = 500
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.25 * (1 - 0.25)/500) = 0.0194

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

Margin of Error, ME = zc * SE
ME = 1.64 * 0.0194
ME = 0.0318

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.25 - 1.64 * 0.0194 , 0.25 + 1.64 * 0.0194)
CI = (0.218 , 0.282)
c. (21.8%, 28.2%)

14)

Rejection Region
This is left tailed test, for α = 0.02
Critical value of z is -2.05
Hence reject H0 if z < -2.05

Test statistic,
z = (xbar - mu)/(sigma/sqrt(n))
z = (2750 - 2900)/(650/sqrt(200))
z = -3.26

a. The test statistic is -3.26 and the critical value is -2.05, hence we reject the null hypothesis.