Question

A major expense in the process of recycling newspapers is the collection of papers from homes. A financial analyst for a firm that collects newspapers for recycling has determined that the firm will make a profit if the average weekly newspaper collection from each household exceeds 2 pounds. In a study to determine the feasibility of a recycling plant, a random sample of 50 households was drawn and the weekly weight of newspapers discarded for recycling was measured. The sample yielded an average weight of 2.18 pounds with a standard deviation of 0.981. At the 5% significance level, do the data provide enough evidence for the analyst to conclude that a recycling plant would be profitable?

Answer 1

The data provides enough evidence for the analyst to conclude that a recycling plant would be profitable.

Step-by-step explanation:

To determine whether the data provide enough evidence for the financial analyst to conclude that a recycling plant would be profitable, we need to conduct a hypothesis test. The null hypothesis (H0) is that the average weekly newspaper collection from each household is less than or equal to 2 pounds. The alternative hypothesis (Ha) is that the average weekly newspaper collection from each household exceeds 2 pounds.

Given that the sample size is 50, the sample mean is 2.18 pounds, and the sample standard deviation is 0.981 pounds, we can perform a t-test. The test statistic is calculated as:

t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))

Using the given values, we get:

t = (2.18 - 2) / (0.981 / sqrt(50)) = 3.272

At a 5% significance level (alpha = 0.05), the critical t-value for a one-tailed test with 49 degrees of freedom is approximately 1.677. Since the calculated t-value (3.272) is greater than the critical t-value, we reject the null hypothesis and conclude that there is enough evidence to support the analyst's claim that a recycling plant would be profitable.

Answer 2

To decide if the recycling plant would be profitable, a one-sample t-test is used to compare the sample mean weight of newspapers collected against the hypothesized mean. The null hypothesis is that the mean weight is less or equal to 2 pounds; we reject this if the t-statistic exceeds the critical value at the 5% significance level, indicating profitability.

Step-by-step explanation:

The question asked requires conducting a hypothesis test to determine if the recycling plant would be profitable by comparing the sample mean of the weekly newspaper collection weight against the hypothesized mean. In this case, a one-sample t-test would be appropriate as we have a sample mean (2.18 pounds), sample standard deviation (0.981), and sample size (50), and we want to test the hypothesis that the true mean weight exceeds 2 pounds.

At the 5% significance level, we first must state our null hypothesis, H0: μ ≤ 2 pounds (the mean is less than or equal to 2 pounds, which suggests the recycling plant would not be profitable), against our alternative hypothesis, H1: μ > 2 pounds (the mean is greater than 2 pounds, which suggests profitability). To conclude whether there is enough evidence to reject the null hypothesis, we would calculate the t-statistic and compare it to the critical value from the t-distribution table. If the t-statistic is greater than the critical value, we reject the null hypothesis, indicating sufficient evidence that the recycling plant will be profitable.

However, the actual calculation of the t-statistic and the decision-making process require additional statistical analysis which is not fully detailed in your question. Calculations and inferences should be done following statistical testing methodologies such as computing the t-statistic and identifying the critical t-value based on the degrees of freedom (n-1) and the significance level.