Question

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

2. the lower limit of two-sided confidence interval for population mean SAT score

c = confidence level = 80% = 0.80

n = sample size = 1000

s = sample standard deviation = 40

The formula of a lower limit of the confidence interval is

where - critical value at a given confidence level.

Degrees of freedom = n - 1 = 1000 - 1 = 999

alpha = 1 - c = 1 - 0.80 = 0.20

t critical value using excel = tinv(alpha, degrees of freedom) = tinv(0.20, 999) = 1.282

The lower limit of 80% confidence interval for the population mean is 578.378384.

3. Claim: The average daily loss from those illegally riding without tickets is at least (greater or equal) $250 that is

n = sample size = 81

s = sample standard deviation = 15

a. alpha = 0.02

The null and alternative hypothesis are,

Test statistics formula

Critical value:

The test is left tailed test since the alternative hypothesis contains the less-than sign.

alpha = 0.02, degrees of freedom = n - 1 = 80

The t critical value using excel is t.inv(0.02,80) = -2.09

Decision rule: |t critical value| (2.09) is greater than |t test statistics| (1.2) so fail to reject the null hypothesis.

Conclusion: Fail to reject the null hypothesis that is there is sufficient evidence to support the claim that the average is at least 250.

b. 90% Confidence interval for mean losses.

Formula

alpha = 1 - 0.90 = 0.10

Using excel the t critical value is =tinv(0.10, 80) = 1.664

The 90% confidence interval for the mean is (245.2266667, 250.7733333)

4. Claim: The average cost of tuition and fees at a 4-year college is less than $5700 that is

n = sample size = 49

s = sample standard deviation = 659

a. The null and alternative hypothesis:

Test statistics:

Alpha = 0.05

Degrees of freedom = n - 1 = 48

Using excel the t critical value is =tinv(0.05, 48) = -1.677

Decision rule: |t critical value| (1.677) is greater than |t test statistics| (0.531) so fail to reject the null hypothesis.

Conclusion: There is no sufficient evidence to support the claim that the average is less than 5700.

b. 90% confidence interval for mean

formula

alpha = 1 - 0.90 = 0.10

t critical value using excel is =tinv(alpha, degrees of freedom) = tinv(0.10, 48) = 1.677

The 90% confidence interval for the population mean is (5592.122429, 5907.877571)

5. Two-sided 90% confidence interval for the population mean.

n = sample size = 144

s = sample standard deviation = 5

Formula of the confidence interval

alpha = 1 - c = 1 - 0.90 = 0.10

Degrees of freedom = n - 1 = 143

t critical value using excel = tinv(0.10, 143) = 1.656

The confidence interval for population mean is (249.31, 250.69)