Answer to Question #251393 in Statistics and Probability for Carrie

Question

In a survey taken 10 years ago, it was found that 10% of customers of a supermarket brought along

their own shopping bags. A recent survey aimed to prove that the current percentage of customers

bringing along their own shopping bags is different from 10%. In the survey, it was found that 92 of

the 1 000 customers surveyed brought along their own shopping bags. We want to test the claim

that the current percentage is not 10%, at the 5% significance level.

(a) State the appropriate null and alternative hypothesis. (2)

(b) State and calculate the appropriate test statistic. (8)

(c) Determine the critical value of the test or the p–value of the test. (4)

(d) State whether or not you reject the null hypothesis, giving the reason. (3)

(e) Draw an appropriate conclusion. (3)

Answer 1

(a) H₀: Current % of customers bringing along their own shopping bags is not different from 10%

H₁: Current % of customers bringing along their own shopping bags is different from 10%

$H_0: p_0 = 0.1 \\ H_1: p_0 ≠ 0.1$

(b) Test-statistic: Z-test for one population proportion

$Z = \frac{\hat{p} -p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \\ Z = \frac{0.092 -0.1}{\sqrt{\frac{0.1(1-0.1)}{1000}}} \\ Z = -0.843$

(c) Critical value at 5% level of significance for two-tailed test

Z_c = 1.96

Rejection region |Z|>1.96

(d) Since it is observed that |Z| = 0.843 < Z_c = 1.96

It is then concluded that null hypothesis is not rejected.

(e) Conclusion: The null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the current % of customers bringing along their own shopping bugs is different from 10% at 5% level of significance.