Homework 2: The First Three Problems Are Designed To Be Comp

Homework 2the First Three Problems Are Designed To Be Completed By Ha

In many areas, newspaper carriers deliver newspapers via automobile due to the length of the routes. In a random sample of 28 newspaper carriers using an automobile, the sample mean route length was 16.7 miles, and the sample standard deviation was 3.4 miles. Assuming the route length is reasonably normally distributed, find the 95% confidence interval for the true mean route length for newspaper carriers who use their automobile to deliver newspapers.

Residential mailboxes in Des Moines, IA should be installed so that the bottom of the mailbox is 42 inches above the ground. A random sample of 75 mailboxes was selected and the height of each mailbox was measured, resulting in a sample mean of 43.22 inches and a sample standard deviation of 7.6 inches. Is there evidence to suggest that the mailboxes in Des Moines, IA are not being installed as required? Conduct a hypothesis test by calculating a p-value, with a significance level α = 0.05.

A 42-inch large screen plasma TV is required to consume no more than 350 Watts of power to receive Energy Star certification. A random sample of 7 TVs with Energy Star certification was selected, with a sample mean power consumption of 353.8 Watts and a standard deviation of 5.6 Watts. Is there evidence that the TVs are not meeting the Energy Star certification standard? Perform a significance test at α = 0.05, assuming the distribution is reasonably normal.

Paper For Above instruction

The following analysis addresses the three primary statistical investigations based on the provided data. The first considers the estimation of the average route length for newspaper carriers, the second examines whether mailboxes are installed to comply with specified standards, and the third evaluates whether plasma TVs meet Energy Star certification power consumption limits.

Problem 1: Confidence Interval for Mean Route Length

Given a sample size (n) of 28 newspaper carriers utilizing automobiles, with a sample mean (x̄) of 16.7 miles and a standard deviation (s) of 3.4 miles, we calculate a 95% confidence interval for the true mean. Since the sample size exceeds 30 and the underlying distribution is assumed normal, the t-distribution is appropriate. The degrees of freedom (df) is n-1 = 27.

The critical t-value (t*) for 95% confidence and 27 degrees of freedom is approximately 2.052 (from t-distribution tables or statistical software). The confidence interval is computed as:

CI = x̄ ± t (s/√n) = 16.7 ± 2.052 * (3.4/√28)

Calculating the margin of error:

ME = 2.052 (3.4/√28) ≈ 2.052 (3.4/5.291) ≈ 2.052 * 0.642 ≈ 1.319

Therefore, the 95% confidence interval is approximately:

(16.7 - 1.319, 16.7 + 1.319) = (15.381, 18.019)

This interval suggests we are 95% confident that the true mean route length for newspaper carriers using automobiles falls between approximately 15.38 and 18.02 miles.

Problem 2: Hypothesis Test on Mailbox Heights

The null hypothesis (H0) states that the population mean height μ is equal to the required standard (42 inches), i.e., H0: μ = 42. The alternative hypothesis (H1) suggests that the mean height differs from this standard: H1: μ ≠ 42. Since the sample of 75 mailboxes has a mean of 43.22 inches and a standard deviation of 7.6 inches, a two-tailed t-test is appropriate.

The test statistic (t) is calculated as:

t = (x̄ - μ0) / (s/√n) = (43.22 - 42) / (7.6/√75) ≈ 1.22 / (7.6/8.660) ≈ 1.22 / 0.878 ≈ 1.389

With df = 74, the p-value corresponding to t = 1.389 for a two-tailed test can be obtained from t-distribution tables or software, yielding approximately 0.168.

Since p = 0.168 > α = 0.05, we fail to reject H0. There is insufficient evidence to conclude that the average mailbox height differs from the required 42 inches in Des Moines, IA.

Problem 3: Test of Power Consumption for Televisions

The null hypothesis (H0): μ ≤ 350 Watts (TVs meet Energy Star standard). The alternative hypothesis (H1): μ > 350 Watts (TVs do not meet the standard). With a sample size of 7, mean of 353.8 Watts, and standard deviation of 5.6 Watts, perform a one-sample t-test.

The test statistic:

t = (x̄ - μ0) / (s/√n) = (353.8 - 350) / (5.6/√7) ≈ 3.8 / (5.6/2.6458) ≈ 3.8 / 2.119 ≈ 1.79

Degrees of freedom: df = 6. Using t-distribution tables or software, the p-value for t = 1.79 with 6 df in a one-tailed test is approximately 0.065.

Since p ≈ 0.065 > α = 0.05, we fail to reject H0 at the 5% significance level. There is insufficient evidence to conclude that the TVs are not meeting the Energy Star power consumption standard.

These analyses leverage standard statistical methods assuming normality and appropriate use of t-distributions. The conclusions suggest the estimated means and standard deviations provide relevant insights into the populations studied, with all tests indicating insufficient evidence to reject null hypotheses at the specified significance levels.

References

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