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Chrysler Corporation agrees to allow you to randomly select 40 of its new Dodge Caravans to test the highway mileage. Chrysler claims that the caravans get 28mpg on the highway. Your results show a mean of 26.7 and a standard deviation of 4.2. You support Chrysler's claim.
1. Show why you support Chrysler's claim by listing the p-value from your output. Test the variability of the miles per gallon on the highway. from futher questioning of Chrysler's quality control engineers, you find they are claiming a standard deviation of 2.1
2. Test the claim about standard deviation
3. What action should Chrysler take to remedy complaints.
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Tests hypothesis about mean and standard deviation using z test and chi square test.
Please see attached file
Chrysler Corporation agrees to allow you to randomly select 40 of its new Dodge Caravans to test the highway mileage. Chrysler claims that the caravans get 28 mpg on the highway. Your results show a mean of 26.7 and a standard deviation of 4.2. You support Chrysler's claim.
1. Show why you support Chrysler's claim by listing the p-value from your output.
Note: Since significance level for testing the hypothesis is not mentioned we can assume it to be 0.05
Hypothesized mean= 28 mpg
Standard deviation= 4.2 mpg
Sample mean= 26.7 mpg
Sample size= 40
Significance level= 0.05 0r 5%
Population standard deviation (Known, Not Known)= Not Known
Null Hypothesis: Ho: M = 28 (:Mean is 28 mpg)
Alternative Hypothesis: H1: M < 28 :( Mean is less than 28 mpg )
Significance level=alpha (a) = 0.05 or 5%
No of tails= 1 ( Left tail )
This is a 1 tailed ( Left tail ) test because we are testing that M < 28
2) Decision rule
sample size=n= 40
Since either sample size is more than 30 , use normal distribution
Thus we use z distribution
Z at the 0.05 level of significance 1 tailed test = 1.6449
z critical = -1.6449
if sample statistic is ...
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