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Down the toilet A company that makes hotel toilets claims that its new pressure-assisted toilet reduces the average amount of water used by more than 0.5 gallon per flush when compared to its current model. To test this claim, the company randomly selects 30 toilets of each type and measures the amount of water that is used when each toilet is flushed once. For the current-model toilets, the mean amount of water used is 1.64 gal with a standard deviation of 0.29 gal. For the new toilets, the mean amount of water used is 1.09 gal with a standard deviation of 0.18 gal. (a) Carry out an appropriate significance test. What conclusion would you draw? (Note that the null hypothesis is not \(H_{0} : \mu_{1}-\mu_{2}=0\) ) (b) Based on your conclusion in part (a), could you have made a Type I error or a Type II error? Justify your answer.

Step by step solution

01

Define the Hypotheses

For part (a), we are conducting a significance test to compare two means. Let's define- **Null Hypothesis**: \( H_0: \mu_1 - \mu_2 = 0.5 \) gallons, where \( \mu_1 \) is the mean water usage for the current model, and \( \mu_2 \) is the mean for the new model.- **Alternative Hypothesis**: \( H_a: \mu_1 - \mu_2 > 0.5 \) gallons, indicating the new toilets use significantly less water, supporting the company's claim.

02

Choose the Significance Level and Test Statistic

Assume a significance level \( \alpha = 0.05 \). We'll use a two-sample t-test for the difference of means. The test statistic is calculated using:\[ t = \frac{(\bar{x}_1 - \bar{x}_2) - d_0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]where \( \bar{x}_1 = 1.64 \), \( \bar{x}_2 = 1.09 \), \( d_0 = 0.5 \), \( s_1 = 0.29 \), \( s_2 = 0.18 \), and both \( n_1 = n_2 = 30 \).

03

Calculate the Test Statistic

First, compute the standard error:\[ SE = \sqrt{\frac{0.29^2}{30} + \frac{0.18^2}{30}} \approx 0.065 \]Now, compute the test statistic:\[ t = \frac{(1.64 - 1.09) - 0.5}{0.065} = \frac{0.05}{0.065} \approx 0.769 \]

04

Determine the Critical Value and Make the Decision

For \( df = 58 \) (using \( n_1 + n_2 - 2 \)), the critical value of \( t \) at \( \alpha = 0.05 \) for a one-tailed test is approximately 1.671. Since our calculated \( t = 0.769 \) is less than 1.671, we fail to reject the null hypothesis.

05

Evaluate Possible Errors

Since we failed to reject the null hypothesis, there is no strong evidence to support the claim that the new toilets reduce water usage by more than 0.5 gallons per flush. This could mean a Type II error might have occurred: we fail to detect a true effect (if it exists). A Type I error, rejecting a true null hypothesis, could not have occurred because we did not reject it.

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Key Concepts

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

Statistical Significance

Statistical significance is a critical concept in hypothesis testing that helps us decide whether the results of an experiment are meaningful. When we conduct a hypothesis test, we aim to determine whether the observed differences or relationships in our data are likely to be genuine or merely occurred by random chance.
This is often done by calculating the p-value. A p-value tells us how likely it is to see results at least as extreme as those observed in the data, assuming the null hypothesis is true.

• A low p-value (usually less than a chosen significance level, such as 0.05) suggests the results are statistically significant, meaning it's unlikely that the observed effect is due to chance alone.
• Conversely, a high p-value indicates the results are not statistically significant, and we're likely seeing random error in our measurements.

In the exercise about the toilets, the calculated p-value did not provide strong evidence against the null hypothesis, which means the finding was not statistically significant.

Type I Error

Type I error, also known as a false positive, occurs when we incorrectly reject the null hypothesis. In simpler terms, we conclude that there is an effect or difference when, in fact, there isn't one.
It's important to control the probability of a Type I error because claiming a false effect could lead to incorrect or misleading conclusions.

• The probability of making a Type I error is represented by the significance level \( \alpha \), often set at 0.05. This means we're willing to risk a 5% chance of incorrectly rejecting a true null hypothesis.
• If our p-value is lower than \( \alpha \), we reject the null hypothesis, with the understanding that we may risk a Type I error.

In the exercise, since we did not reject the null hypothesis, a Type I error could not have occurred because there was no claim made about finding a significant difference.

Type II Error

Type II error, or a false negative, occurs when we fail to reject a false null hypothesis. Essentially, it means missing an effect or difference that actually exists.
Understanding Type II errors is crucial because it affects our ability to detect true differences or effects in data.

• One way to reduce the risk of a Type II error is by ensuring a sufficiently large sample size, which enhances the test's power to detect actual effects.
• The probability of making a Type II error is denoted by \( \beta \), and power is calculated as \( 1 - \beta \). Higher power means a lower chance of committing a Type II error.

In the toilet exercise, we failed to reject the null hypothesis. This means we may have committed a Type II error by not finding evidence for the true difference in water usage between the two toilet models.

Two-sample t-test

The two-sample t-test is a statistical method used to compare the means of two groups to determine if they differ significantly from each other. This test assesses if the difference between the means of two samples is statistically significant.
The two-sample t-test is useful in scenarios where we want to evaluate the effect of an intervention or the difference between two groups.

• For example, in the toilet exercise, we used a two-sample t-test to compare the mean water usage of the new versus the current toilet models.
• The test involves calculating the test statistic (t-value) and comparing it to a critical value from the t-distribution based on the chosen significance level and degrees of freedom.

To perform this test, certain assumptions must be met: both samples should be independent, normally distributed, and have equal variances. In this exercise, the test showed no significant difference between the two toilet models within the context of the alternately specified null hypothesis.