91Ó°ÊÓ

As mentioned in Exercise \(10.26\), a town that recently started a single-stream recycling program provided 60 -gallon recycling bins to 25 randomly selected households and 75 -gallon recycling bins to 22 randomly selected households. The average total volumes of recycling over a 10 -week period were 382 and 415 gallons for the two groups, respectively, with standard deviations of \(52.5\) and \(43.8\) gallons, respectively. Suppose that the standard deviations for the two populations are not equal. a. Construct a \(98 \%\) confidence interval for the difference in the mean volumes of 10 -week recyclying 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 10 -week recycling volume of all households that have 75 -gallon containers? c. Suppose that the sample standard deviations were \(59.3\) and \(33.8\) gallons, respectively. Redo parts a and b. Discuss any changes in the results.

Step by step solution

01

Calculate Confidence Interval

Using the formula for confidence interval when the variances are unequal: \(CI = (\bar{X1} - \bar{X2}) \pm t_{\alpha/2} * sqrt((s1^2 / n1) + (s2^2 / n2))\) where \(\bar{X1}\) and \(\bar{X2}\) are the sample means, \(s1\) and \(s2\) are the sample standard deviations, \(n1\) and \(n2\) are the sample sizes and \(t_{\alpha/2}\) is the t critical value for \(\alpha/2\) degrees of freedom. Filling in given values for 60-gallon and 75-gallon groups, we have: \(CI = (382 - 415) \pm t_{0.01}* sqrt((52.5^2 / 25) + (43.8^2 / 22))\). Now, after getting the value of \(t_{0.01}\) from t-table, calculate the CI.

02

Perform Hypothesis Test

We'll use a t-test for two means with unknown and unequal standard deviations. The null hypothesis will be \(H0: µ1 = µ2\) that there is no difference between the means of the two groups, and the alternative hypothesis will be \(Ha: µ1 ≠ µ2\) that there is a difference. We calculate the test statistic by: \(t = (\bar{X1} - \bar{X2}) / sqrt((s1^2 / n1) + (s2^2 / n2))\). If the absolute value of t is greater than \(t_{\alpha/2}\) we reject the null hypothesis, otherwise we fail to reject it.

03

Redo Parts (a) and (b) with Different Standard Deviations

Do the same calculations as in step 1, but replace the sample standard deviations with the new standard deviations, i.e. \(s1 = 59.3\) and \(s2=33.8\). Compute the new confidence interval and t value. Then, compare these new results with the ones obtained from step 1 and 2

04

Interpretation

Present the results of the calculations in a way that answers the initial exercise. Make sure to answer whether the difference in container sizes makes a significant impact on the amount of recycling, according to the statistics and considering the changes in standard deviation.

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Key Concepts

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

Understanding Hypothesis Testing

Hypothesis testing is a statistical procedure used to determine if there is enough evidence to reject a hypothesis about a population parameter. In our recycling bin example, we want to investigate whether households with different bin sizes have different recycling habits.

To do this, we start with a null hypothesis, denoted as \( H_0 \), which assumes there is no difference in mean recycling volumes between the two groups. The alternative hypothesis, \( H_a \), suggests there is indeed a difference.

• **Null Hypothesis (\(H_0\))**: \( \mu_1 = \mu_2 \)
• **Alternative Hypothesis (\(H_a\))**: \( \mu_1 eq \mu_2 \)

In hypothesis testing, we often set a significance level, also known as alpha (\( \alpha \)), which is the probability of rejecting the null hypothesis when it is actually true. In part (b) of the exercise, this level is set at 0.02 or 2%.By conducting hypothesis testing, we compare the calculated test statistic against a critical value from the statistical distribution table. If the test statistic is greater than the critical value, we reject the null hypothesis in favor of the alternative.

The Role of Standard Deviation

Standard deviation is a statistical measure that describes the dispersion or spread of data points in a data set. In simpler terms, it tells us how much the values in the data set deviate from the mean (average).

In our recycling example, the standard deviations for each group give us an idea of the variability in recycling amounts for households with different bin sizes. The standard deviations provided are 52.5 gallons for the 60-gallon bin group, and 43.8 gallons for the 75-gallon bin group.

• A **larger standard deviation** suggests more variability in the data set.
• A **smaller standard deviation** means the data points are closer to the mean.

These values are crucial for calculating the confidence interval and performing the t-test, as they impact the precision of our estimates of the population means.

Conducting a t-Test

The t-test is a statistical test used to compare the means of two groups to determine if they are significantly different from each other. It is particularly useful when dealing with smaller sample sizes and unknown population variances.

In the exercise, we apply a t-test for two independent samples with unequal standard deviations. This scenario often arises when comparing different groups. We compute the t-test statistic using:\[ t = \frac{\bar{X_1} - \bar{X_2}}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]

Where:

• \(\bar{X_1}\) and \(\bar{X_2}\) are the sample means,
• \(s_1\) and \(s_2\) are the sample standard deviations,
• \(n_1\) and \(n_2\) are the sample sizes.

With the critical value found in the t-distribution table, we compare it to our calculated t statistic. This test helps us determine if we can reject the null hypothesis, thereby concluding if bin sizes significantly affect recycling volumes.

Importance of the Sample Mean

The sample mean, often referred to as the average, is a central value for a set of data, calculated by dividing the sum of the values by the number of values. It provides a quick snapshot of the data’s central tendency.

In our recycling context, the sample mean represents the average recycling volume for each group of households — those with 60-gallon bins and those with 75-gallon bins. The sample means are 382 and 415 gallons, respectively.

• Understanding sample means helps us assess overall data trends and make predictions about larger populations.
• The sample mean serves as a basis for further statistical analysis, such as calculating confidence intervals and performing hypothesis tests.

By comparing the sample means between two groups, we form a foundational part of our analysis, ultimately driving decisions on whether to accept or reject hypotheses about differences between the groups.