Question

9. A company wants to know if the proportion of consumers in Houston that recognizes its brand name is higher than the proportion of consumers in Austin that recognizes its brand name. In a random sample of consumers in Houston, 446 out of 705 consumers recognized the company’s brand name. In a random sample of consumers in Austin, 411 out of 715 consumers recognized the company’s brand name. Make Houston group 1 and make Austin group 2.

(a) What is the null hypothesis and what is the alternative hypothesis?

(b) What is the sample proportion for group 1? (round to 5 digits after the decimal place)

(c) What is the sample proportion for group 2? (round to 5 digits after the decimal place)

(d) What is the pooled estimator for p? (round to 5 digits after the decimal place)

(e) What is the standard error for the difference in the sample proportions? (Use ˜σp1−p2 and round to 5 digits after the decimal place.)

(f) What is the value of the test statistic? (Round to 2 digits after the decimal place.)

(g) What is the p-value of the test? (Round to 3 digits after the decimal place.)

(h) Do we reject or not reject the null hypothesis at the .05 level of significance? Reject Not reject

(i) Can we interpret the difference in the population proportions as a causal effect? Yes, it has a causal interpretation. or No, it does not have a causal interpretation.

Answer 1

solution:

let group 1 for Houston

sample size = 705

number of customer who recognize brand = = 446

let group 2 for Austin

sample size = 715

number of customer who recognize brand = = 411

a) Null and Alternative Hypotheses

b)

sample proportion for group 1 =

c) sample proportion for group 2 =

d)

The value of the pooled proportion is computed as =

e) standard error for the difference in the sample proportion

f) test statistics:

g) p value = 1 - value of z to the left of 2.23 from the z table

p value = 0.013

h) since p value 0.013 < 0.05, so rejecting the null hypothesis.

i) conclusion:

there is sufficient evidence that the proportion of Hauston is higher than the Austin

This corresponds to a left-tailed test, for which a z-test for two population proportions needs to be conducted.

test statistics:

significance level = 0.05

p value = value of z to the left of -2.46 = 0.0069

since p value 0.0069 < 0.05

so null hypothesis is rejected

conclusion:

there is enough evidence to support the claim that the truck violate the rule at a higher rate than the car owner.

confidence interv