Testing of hypothesis

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Please see attached file:
1 23. Microwave Repair Costs A microwave oven repairer says that the mean repair cost for damaged microwave ovens is less than $100. You work for the repairer and want to test this claim. You find that a random sample of five microwave ovens has a mean repair cost of $75 and a standard deviation of $12.50. At a = 0.01,do you have enough evidence to support the repairer's claim? (Adapted from Consumer Reports)

Data
Hypothesized mean= $100
Standard deviation= $12.50
Sample mean= $75
Sample size= 5
Significance level= 0.01 0r 1%
Population standard deviation (Known, Not Known)= Not Known

1) Hypothesis
Null Hypothesis: Ho: M = 100 (:Mean is 100 )
Alternative Hypothesis: H1: M < 100 :( Mean is less than 100 )
Significance level=alpha (a) = 0.01 or 1%
No of tails= 1 ( Left tail )
This is a 1 tailed ( Left tail ) test because we are testing that M < 100

2) Decision rule
sample size=n= 5
Since sample size 5 is less than 30 and population standard deviation is 'Not Known' use t distribution
Thus we use t distribution
t at the 0.01 level of significance and 4 degrees of freedom (=n-1) and 1 tailed test= 3.7469
t critical = -3.7469

if sample statistic is <-3.7469 Reject Null Hypothesis, else Accept Null Hypothesis
Alternatively,
if p value is less than the significance level (= 0.01 ) Reject Null Hypothesis, else Accept Null Hypothesis

3) Calculation of sample statistics

Hypothesized Mean=M = 100 0
Standard deviation =s= 12.5 0
sample size=n= 5
sx=standard error of mean=s/square root of n= 5.5902 = ( 12.5 /square root of 5)
sample mean= 75

t=(sample mean-M )/sx= -4.4721 =(75-100)/5.5902

4) Compare sample statistic with critical value

test statistic= -4.4721
t critical = -3.7469

therefore, sample statistic is outside acceptance region

5) Decision

Reject Null Hypothesis :( Mean is less than 100 )

Alternatively, calculate p value
Prob-value corresponding to t = -4.4721 is 0.55% which is less than the significance level of 1%
Reject Null Hypothesis since probability value is less than the level of significance
:( Mean is less than 100 )

2 37. Gas Mileage A car company says that the mean gas mileage for its luxury sedan is at least 21 miles per gallon (mpg).You believe the claim is incorrect and find that a random sample of five cars has a mean gas mileage of 19 mpg and a standard deviation of 4 mpg. Assume the gas mileage of all of the company's luxury sedans is normally distributed. At a = 0.05,test the company's claim. (Adapted from Consumer Reports)

Data
Hypothesized mean= 21 mpg
Standard deviation= 4 mpg
Sample mean= 19 mpg
Sample size= 5
Significance level= 0.05 0r 5%
Population standard deviation (Known, Not Known)= Not Known

1) Hypothesis
Null Hypothesis: Ho: M = 21 (:Mean is at least 21 mpg)
Alternative Hypothesis: H1: M < 21 :( Mean is less than 21 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 < 21

2) Decision rule
sample size=n= 5
Since sample size 5 is less than 30 and population standard deviation is 'Not Known' use t distribution
Thus we use t distribution
t at the 0.05 level of significance and 4 degrees of freedom (=n-1) and 1 tailed test= 2.1318
t critical = -2.1318

if sample statistic is <-2.1318 Reject Null Hypothesis, else Accept Null Hypothesis
Alternatively,
if p ...