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Use the data from problem #1. Assume a probability normal probability plot suggest the data could come from a population that1. The data below represent the repair cost for a low-impact collision in a simple random sample of mini- and micro- vehicles

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Answer #1

Calculations for the mean and standard deviation are given below

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(a) One sample t test for mean.

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(b) The Hypothesis:

H0: \mu = 1200

Ha: \mu > 1200

This is a right tailed (one tailed test)

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(c) The Test Statistic:

t = \frac{\bar{x}-\mu }{\frac{s}{\sqrt{n}}} = \frac{1756.5-1200}{\frac{1007.4542}{\sqrt{10}}} = 1.75

P value = 0.057. Therefore the p value range is 0.05 < p value < 0.10

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(d) Since p value is not mentioned, we use the defualt level = 0.05.

Since p value is > 0.05, Then fail to reject H0.

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(e) The conclusion is that there is insufficient evidence at the 95% significance level to conclude that the average cost of repairs for mini and micro vehicles who have been in a low impact collision is greater than $1200.

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Please note: In case Alpha is 0.10, the the decision would be to reject H0 as p value would be < Alpha.

The conclusion would be that there is sufficient evidence at the 90% significance level to conclude that the average cost of repairs for mini and micro vehicles who have been in a low impact collision is greater than $1200.

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Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)2.

# X Mean (x - mean)2
1 3148 1756.50 1936272.25
2 1758 1756.50 2.25
3 1071 1756.50 469910.25
4 3345 1756.50 2523332.25
5 743 1756.50 1027182.25
6 2057 1756.50 90300.25
7 663 1756.50 1195742.25
8 2637 1756.50 775280.25
9 773 1756.50 967272.25
10 1370 1756.50 149382.25
Total 17565 9134676.50
n 10
Sum 17565
Average 1756.50
SS(Sum of squares) 9134676.5
Variance = SS/n-1 1014964.056
Std Dev=Sqrt(Variance) 1007.4542
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