Question

CEO of SABA Co. claims that about 90 percent of the people who purchase contact lenses from them experience no difficulty. A consumer agency thinks that less than 90 percent of customers are happy with SABA’s product, therefore, from the company files, the agency takes a random sample of 400 people who have purchased contact lenses and finds that 340 have experienced no difficulty. Test agency’s claim at the 0.05 level of significance. 1) What are the alternative and null hypotheses? 2) What is the Level of Significance for this test? 3) What is the computed Z value for this test? 4) What is the computed p-value for this test? 5) What is the decision? a. Yes. The agency’s claim is valid. b. No. The agency’s claim is not valid. c. both a and b are correct, depending on the total number of customers. d. None of the above. 6) What is the 99% confidence interval for percentage of customers who are happy with SABA CO’s contact lenses?

Answer 1

1) H₀: P = 0.9

    H₁: P < 0.9

2) The level of significance is 0.05.

3) $\widehat p$ = 340/400 = 0.85

The test statistic z = ( $\widehat p$ - P)/sqrt(P(1 - P)/n)

                              = (0.85 - 0.9)/sqrt(0.9 * 0.1/400)

                              = -3.33

4) P-value = P(Z < -3.33)

                 = 0.0004

5) Since the P-value is less than the significance level (0.0004 < 0.05), so we should reject the null hypothesis.

Option - a) Yes, the agency's claim is valid.

6) At 99% confidence interval the critical value is z_0.005 = 2.58

The 99% confidence interval for population proportion is

$\widehat p$ +/- z_0.005 * sqrt( $\widehat p$ (1 - $\widehat p$ )/n)

= 0.85 +/- 2.58 * sqrt(0.85 * (1 - 0.85)/400)

= 0.85 +/- 0.046

= 0.804, 0.896