Question

A recent study of two vendors of desktop personal computers reported that out of 876 units sold by Brand A, 122 required repair, while out of 758 units sold by Brand B, 92 required repair. Round all numeric answers to 4 decimal places. 1. Calculate the difference in the sample proportion for the two brands of computers, p^BrandA−p^BrandB = .

2. What are the correct hypotheses for conducting a hypothesis test to determine whether the proportion of computers needing repairs is different for the two brands. A. H0:pA−pB=0 , HA:pA−pB≠0 B. H0:pA−pB=0, HA:pA−pB<0 C. H0:pA−pB=0, HA:pA−pB>0

3. Calculate the pooled estimate of the sample proportion, p^ =

4. Is the success-failure condition met for this scenario? A. Yes B. No

5. Calculate the test statistic for this hypothesis test. =

6. Calculate the p-value for this hypothesis test, p-value = .

7. What is your conclusion using α = 0.1? A. Do not reject H0 B. Reject H0

8. Compute a 90 % confidence interval for the difference p^BrandA−p^BrandB = ( ,

Answer 1

Solution:-

1)

P_A = 122/876 = 0.1393

P_B = 92/758 = 0.1214

2)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P_A = P_B
Alternative hypothesis: P_A P_B

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

3)

p = (p₁ * n₁ + p₂ * n₂) / (n₁ + n₂)
p = 0.13097

4) (A) Yes, the success-failure condition are met for this scenario.
SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.01674

5)
z = (p₁ - p₂) / SE

z = 1.07

where p₁ is the sample proportion in sample 1, where p₂ is the sample proportion in sample 2, n₁ is the size of sample 1, and n₂ is the size of sample 2.

6)

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -1.07 or greater than 1.07.

Thus, the P-value = 0.2846

Interpret results. Since the P-value (0.2846) is greater than the significance level (0.10), we cannot rejectt the null hypothesis.

7) A. Do not reject H₀

8) 90 % confidence interval for the difference pA−pB = C.I = ( - 0.198, 0.233)

C.I = 0.0179 + 1.645 × 0.13097

C.I = 0.0179 + 0.21545

C.I = ( - 0.198, 0.233)