# Test of Hypothesis -Testing Gas Mileage Claims

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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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

Data

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

1) Hypothesis

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 ...

#### Solution Summary

Tests hypothesis about mean and standard deviation using z test and chi square test.