Math Homework for Statistics

Question

Confidence Interval 

Confidence Level

90%

95%

99%

Critical Value z*

1.645

1.96

2.576

5. Two-hundred and fifty randomly selected people

are surveyed to determine if they own a tablet. 

Of the 250 surveyed, 98 reported owning a

tablet. Find the 90 % confidence interval for the

people who own tablets.

6. You filled your car 100 times and found the mean

was 23.4 mpg with the standard deviation of .9

miles per gallon. Compute the 90% confidence

interval for your miles per gallon.

Do the results of problem #5 support that 42% of

people own a tablets? Why or why not.

(Hint use the results from #6 to determine) A

fuel-efficient four-wheel drive vehicle gets 24

mpg. Does this interval provide evidence that

your truck is fuel-efficient? Why or why not.

Answer 1

5. n=250, z*=1.645 based on the z-score for 90% CI

p̂=98/250=0.392

CI for proportions: p̂ ± z*(SE) where SE = sqrt((p̂*(1-p̂))/n)

SE = sqrt((0.392*(1-0.392))/250) = 0.031

CI: 0.392 ± 1.645(0.031) = 0.392 ± 0.051 -> (0.341, 0.443)

6. x̄=23.4 mpg, σ=0.9 mpg, n=100

CI for known SD: x̄ ± z*(SE) where SE = σ/sqrt(n)

SE = 0.9/sqrt(100) = 0.09

CI: 23.4 ± 1.645(0.09) = 23.4 ± 0.148 -> (23.252, 23.548)

7. Yes, 42% (0.42) is included in the confidence interval for the proportion of people who own tablets of (0.341, 0.443).

8. No, 24 mpg is not included in the confidence interval for the mean mpg of the truck of (23.252, 23.548).