Question

1.Mall security estimates that the average daily per-store theft is no more than $250, but wants to determine the accuracy of this statistic. The company researcher takes a sample of 64 clerks and finds that =$252 and s = $10. a) Test at α = .05 b) Construct a 90% CIE of μ 2.A college wants to know the lower limit of a two-sided 80% confidence interval for SAT scores on the Math section (range equals 200 to 800). They sampled 1000 students and found that the sample mean was 580 and the sample standard deviation was 40. 3.A Rollercoaster's auditors estimate that the average daily loss from those illegally riding without tickets is at least (greater or equal) $250, but wants to determine the accuracy of this statistic. The company researcher takes a random sample of losses over 81 days and finds that = $248 and s = $15. a) Test at α = 0.02. b) Construct a 90% confidence interval of losses Note: a and b are independent. 4. A researcher wishes to test the claim that the average cost of tuition and fees at a 4-year college is less than $5700. She selects a random sample of 49 colleges and finds the mean to be $5750. The sample standard deviation is $659. a) Is there evidence to support the claim at α = .05? b) Construct a 90% confidence interval. 5. A company that manufactures special generators is interested in constructing a two-sided 90% confidence interval for the population mean. They sampled 144 generators and found that the sample mean is 250 hours and the sample standard deviation is 5 hours.

Answer 1

SOLUTION 1 a] NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS Ha:

n=64 , mean= $252 and s= $10

Test statistic t= xbar-mean/s/sqrt(n)

t= 252-250/10/sqrt(10)

t= 2/10/3.16

t= 2/3.16

t= 0.633

Degrees of freedom= n-1=64-1=63

P value= ( by using TDIST we get the P value )

Since P value GREATER than 0.05 therefore NOT SIGNIFICANT

Decision: DO NOT REJECT NULL HYPOTHESIS H0.

Conclusion: We DO NOT have sufficient evidence to conclude that the Population mean is different from $250.

SOLUTION b] sample Mean = $252
t critical = 1.67
sM = √(102/64) = 1.25
μ = M ± t(sM)
μ = 252 ± 1.67*1.25
μ = 252 ± 2.09

90% CI [249.91, 254.09].

You can be 90% confident that the population mean (μ) falls between 249.91 and 254.09.

NOTE: AS PER THE GUIDELINES I HAVE DONE THE FIRST QUESTION PLEASE RE POST THE REST.