Question

A package delivery service wants to compare the proportion of on-time deliveries for two of its major service areas. In City A, 314 out of a random sample of 378 deliveries were on time. A random sample of 280 deliveries in City B showed that 221 were on time.

1. Calculate the difference in the sample proportion for the delivery times in the two cities. ?̂?????−?̂???yB

2. What are the correct hypotheses for conducting a hypothesis test to determine whether the proportion of deliveries that are on time in City A is different from than the proportion in City B? A. ?0:??=??, ??:??≠?? B. ?0:??=?B, ??:??>?B C. ?0:??=pB, ??:??<??

3. Calculate the pooled estimate of the sample proportion. ?̂=

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

5. Calculate the test statistic for this hypothesis test. z = 6.

Calculate the p-value for this hypothesis test. p-value = 7.

Based on the p-value, we have: A. little evidence B. extremely strong evidence C. very strong evidence D. some evidence E. strong evidence that the null model is not a good fit for our observed data.

8. Compute a 95% confidence interval for the difference ?̂?????−?̂?????

need help with #6,7,8

Answer 1

1. For City A :

n1 = 378

x1 = 314

p̂1 = x1/n1 = 0.8307

For City B :

n2 = 280

x2 = 221

p̂2 = x2/n2 = 0.7893

difference = (p̂1 - p̂2) = (0.8307 - 0.7893) = 0.0414

2.

Null and Alternative hypothesis:

Ho : p1 = p2

H1 : p1 ≠ p2

3.

Pooled proportion:

p̄ = (x1+x2)/(n1+n2) = (314+221)/(378+280) = 0.8131

4.

Yes, the success-failure condition met for this scenario

5.

Test statistic:

z = (p̂1 - p̂2)/√ [p̄*(1-p̄)*(1/n1+1/n2)] = (0.8307 - 0.7893)/√[0.8131*0.1869*(1/378+1/280)] = 1.3469

6.

p-value = 2*(1-NORM.S.DIST(ABS(1.3469), 1)) = 0.1780

7.

Decision:

p-value > α, Do not reject the null hypothesis

A. little evidence

8.

95% Confidence interval for the difference:

At α = 0.05, two tailed critical value, z_c = NORM.S.INV(0.05/2) = 1.960

Lower Bound = (p̂1 - p̂2) - z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ] = (0.8307 - 0.7893) - 1.96*√[(0.8307*0.1693/378) + (0.7893*0.2107/280)] = -0.0195

Upper Bound = (p̂1 - p̂2) + z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ] = (0.8307 - 0.7893) + 1.96*√[(0.8307*0.1693/378) + (0.7893*0.2107/280)] = 0.1023

-0.0195 < p1 -p2 < 0.1023