Question

1. The Allstate Insurance Company determined the average number of years between accidents for drivers in a large number of U.S. cities. The ‘Accidents’ sheet in the ‘Lab12 Chp 10n11 S20’ spreadsheet contains the results for 32 randomly selected cities east of the Mississippi River and 32 randomly selected cities west of the Mississippi River.

(a) Are the samples independent or paired?

(b) Consider constructing a 90% confidence interval for the difference in average number of years between accidents of western and eastern cities.

               i. Find the point of estimate using proper notation. Show your work including all relevant quantities needed. Round your answer to 2 decimal places.

               ii. Find the margin of error. Show your work including all relevant quantities needed using proper notation. Round your answer to 4 decimal places.

               iii. Construct the confidence interval by filling in the blanks below. Round values to 2 decimal places.

_______________ < _______________ < _______________

               iv. An insurance company executive claims that the average number of years between accidents for western cities is at least 4 months greater than the average for eastern cities. Does the confidence interval contradict this claim? Justify your answer.

(c) Consider the question: can you conclude that the average time between accidents differs for western cities compared to eastern cities at a 25% significance level?

               i. State the null and alternate hypothesis using proper notation.

               ii. Circle the type of critical value(s)/test statistic and corresponding region needed to answer the problem.

                         • z-score or t-score

                         • Identify the degrees of freedom, if not applicable write NA:

                         • right-tailed, left-tailed, or two-tailed

               iii. Find the critical value(s) using proper notation. Round values to 3 decimal places.

               iv. Find the test statistic using proper notation. Show your work including all relevant quantities needed. Round your answer to 4 decimal places.

               v. Compute the P-value of the test statistic. Round your answer to 4 decimal places.

               vi. Can you reject the null hypothesis? Justify your answer.

               vii. Answer the question.

East	West
9.5	9.5
10.2	8.9
11.2	13.4
11.4	12.6
9.8	9.7
10	8.2
11.6	14
10.7	8.1
9.5	10.2
9	9.4
7.2	10.8
11.5	9.5
7.7	8.5
9	9.1
8.9	8.4
8.4	7.6
6.8	7.9
5.3	7.7
8	7.1
8.8	11.9
9	10.2
10	11.9
9.7	7.7
11.6	8
7.9	10
9.9	8.1
10.5	11.5
11.9	8.4
10.9	9.6
8.6	9.5
6.7	9.9
9.6	11.5

Please Help and Thank You!

Answer 1

For West :

∑x = 308.8

∑x² = 3077.14

n1 = 32

Mean , x̅1 = Ʃx/n = 308.8/32 = 9.6500

Standard deviation, s1 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(3077.14-(308.8)²/32)/(32-1)] = 1.7709

For East :

∑x = 300.8

∑x² = 2907.7

n2 = 32

Mean , x̅2 = Ʃx/n = 300.8/32 = 9.4000

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(2907.7-(300.8)²/32)/(32-1)] = 1.6082

--

a) No, the samples are not paired.

b) i) Point estimate = (x̅1 - x̅2) = (9.65 - 9.4) = 0.25

ii) At α = 0.1 and df = n1+n2-2 = 62, two tailed critical value, t-crit = T.INV.2T(0.1, 62) = 1.670

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((32-1)*1.7709² + (32-1)*1.6082²) / (32+32-2) = 2.86129

Margin of error, E = t-crit*√(S²p*(1/n1 +1/n2)) = 1.67*√(2.8613*(1/32 + 1/32)) = 0.7061

iii) 90% Confidence interval :

Lower Bound = (x̅1 - x̅2) - E = (9.65 - 9.4) - 0.7061 = -0.4561

Upper Bound = (x̅1 - x̅2) + E = (9.65 - 9.4) + 0.7061 = 0.9561

-0.4561 < µ1 - µ2 < 0.9561

iv) As the confidence interval do not contain 4 and both values are less than 4. so we reject the null hypothesis.

Yes, The confidence interval contradict this claim

c)

Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 ≠ µ2

df = n1+n2-2 = 62

Critical value :

Two tailed critical value, t crit = T.INV.2T(0.25, 62) = 1.161

Reject Ho if t < -1.161 or if t > 1.161

Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((32-1)*1.7709² + (32-1)*1.6082²) / (32+32-2) = 2.8613

Test statistic:

t = (x̅1 - x̅2 - (µ1 - µ2)) / √(s²p(1/n1 + 1/n2 ) = (9.65 - 9.4 - 4) / √(2.8613*(1/32 + 1/32)) = -8.8677

p-value :

Two tailed p-value = T.DIST.2T(ABS(-8.8677), 62) = 0.0000

Decision:

p-value < α, Reject the null hypothesis

Conclusion:

There is enough evidence to conclude that the average time between accidents differs for western cities compared to eastern cities.