91Ó°ÊÓ

A town that recently started a single-stream recycling program provided 60 -gallon recycling bins to 25 randomly selected houscholds and 75 -gallon recycling bins to 22 randomly selected households. The total volume of recycling over a 10 -week period was measured for each of the households. The average total volumes were 382 and 415 gallons for the households with the \(60-\) and 75 -gallon bins, respectively. The sample standard deviations were \(52.5\) and \(43.8\) gallons, respectively. Assume that the 10 -week total volumes of recycling are approximately normally distributed for both groups and that the population standard deviations are equal. a. Construct a \(98 \%\) confidence interval for the difference in the mean volumes of 10 -week recycling for the households with the \(60-\) and 75 -gallon bins. b. Using a \(2 \%\) significance level, can you conclude that the average 10 -week recycling volume of all households having 60 -gallon containers is different from the average volume of all households that have 75 -gallon containers?

Step by step solution

01

State the hypothesis

The null hypothesis \(H_0\) assumes that there is no difference in the means of the volumes of the two groups: \(H_0: \mu_1 - \mu_2 = 0\). The alternate hypothesis \(H_1\) assumes there is a difference: \(H_1: \mu_1 - \mu_2 \neq 0\) where \(\mu_1\) and \(\mu_2\) represent the mean volumes for the 60-gallon and 75-gallon bins, respectively.

02

Compute the test statistics

Statistical testing for difference of means can be performed with a t-test. The test statistic t can be calculated using the formula \(t = \frac{\bar{x_1} - \bar{x_2} - (\mu_1 - \mu_2)}{s_p \sqrt{1/n_1 + 1/n_2}}\). Here, \(\mu_1\) and \(\mu_2\) are the believed true population means, \(\bar{x_1}\) and \(\bar{x_2}\) are the sample means, \(n_1\) and \(n_2\) are the sample sizes, and \(s_p\) is the pooled standard deviation, which can be calculated as \(s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}\). Here \(s_1\) and \(s_2\) are the sample standard deviations for the 60-gallon and 75-gallon bins, respectively.

03

Compute the confidence interval for the difference in means

The \(98\%\) confidence interval for the difference in means can be found by computing \(\bar{x_1} - \bar{x_2} \pm t_{\alpha/2, n_1 + n_2 - 2} s_p \sqrt{1/n_1 + 1/n_2}\) where \(t_{\alpha/2, n_1+n_2-2}\) is the t-value which cuts off the upper \(\alpha/2\) (here, \(0.01\)) proportion of the area in a t-distribution with \(n_1 + n_2 - 2\) degrees of freedom.

04

Make a decision based on the significance level

Compare the computed t-value with the critical t-value for the 2% significance level to make a decision. If the absolute value of the calculated t-value is greater than the critical value, there is enough evidence to reject the null hypothesis. If this is the case, it can be concluded that the average 10-week recycling volume of all households having 60-gallon containers is different from the average volume of all households that have 75-gallon containers.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval

A confidence interval gives you a range of values that is likely to contain the true difference between the means of two populations. In this exercise, we are constructing a 98% confidence interval for the difference in recycling volumes between households with 60-gallon and 75-gallon bins. The formula used is: \[\bar{x_1} - \bar{x_2} \pm t_{\alpha/2, n_1 + n_2 - 2} s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\]- \(\bar{x_1}\) and \(\bar{x_2}\) are the sample means.- \(t_{\alpha/2, n_1+n_2-2}\) is the critical t-value based on our confidence level, which tells us how far our sample mean is likely to be from the true population mean.- \(s_p\) is the pooled standard deviation.If our interval does not contain 0, we have evidence to suggest there is a true difference in recycling volumes.

Significance Level

The significance level is the probability of rejecting the null hypothesis when it is actually true, commonly denoted by \(\alpha\). In our context, a 2% significance level is used. This means our threshold for declaring a significant result is that there is only a 2% chance of observing such data, assuming the null hypothesis is true. This low significance level indicates we require strong evidence to conclude a difference in means. Using a stricter level minimizes the risk of a Type I error, ensuring our results are due to actual differences and not random chance.

Null Hypothesis

The null hypothesis \(H_0\) is an essential part of hypothesis testing. It posits that there is no effect or difference. For this exercise, it assumes that the mean difference in recycling volumes between the households with 60-gallon and 75-gallon bins is zero: \[H_0: \mu_1 - \mu_2 = 0\]By testing this hypothesis, we can decide together using statistical evidence if the apparent difference is statistically significant or if it's just due to random sampling variability. The process involves comparing the calculated t-value to the critical t-value. If our data provide enough grounds to reject \(H_0\), we accept the alternative hypothesis, asserting a true difference.

t-test

The t-test is used to compare the means of two groups to determine if they are significantly different from each other. In this exercise, we utilized a two-sample t-test, which tests if the difference in means of our two groups is statistically significant. The calculation of the t-test statistic is done using:\[t = \frac{\bar{x_1} - \bar{x_2} - (\mu_1 - \mu_2)}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}\]Here:- \(\bar{x_1}\) and \(\bar{x_2}\) are sample means.- \(s_p\) is the pooled standard deviation, reflecting variance within each sample group.- The test helps to decide if the observed difference in sample means reflects a real difference in population means.To conclude if there's a significant difference, we compare the calculated t-value with the critical t-value from the t-distribution table based on our significance level.