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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 $0.5$a 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$1.64$a gal with a standard deviation of $0.29$gal. For the new toilets, the mean amount of water used is $1.09\mathrm{gal}$with a standard deviation of $0.18\mathrm{gal}$.

(a) Carry out an appropriate significance test. What conclusion would you draw? (Note that the null hypothesis is not$\left.H_0:{渭}_1-{渭}_2=0_{-}\right)$

(b) Based on your conclusion in part (a), could you have made a Type I error or a Type Il error? Justify your answer.

Step by step solution

01

Part (a) Step 1: Given Information

${\stackrel{炉}{x}}_1=1.64,{\stackrel{炉}{x}}_2=1.09$

$s_1=0.29,s_2=0.18$

$n_1=30,n_2=30$

02

Part (a) Step 2: Explanation

Test statistic formula is:

$t=\frac{{\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2}{\sqrt{\frac{x_1^7}{n_1}+\frac{x_2^2}{n_2}}}$

The null and alternative hypotheses for the provided case are:

$H_0:{渭}_1-{渭}_2=0.5$

$H_1:{渭}_1-{渭}_2>0.5$

The test statistic is computed as:

$t=\frac{{\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2-\left({渭}_1-{渭}_2\right)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$

$=\frac{1.64-1.09-(0.5)}{\sqrt{\frac{0.{29}^2}{30}+\frac{0.{18}^2}{30}}}$

$=0.802$

The degree of freedom is calculated as:

$df=\min \left(n_1-1,n_2-1\right)=\min (30-1,30-1)=29$

The p-value:

$P-\text{value}=0.828$

In this case, $P-value=0.828>0.05$

The null hypothesis could not be rejected which is not showing sufficient evidence for the claim at a significant level of $5\backslash \%.$

03

Part (b) Step 1: Given Information

To determine the error that is committed using the result of the above part.

04

Part (b) Step 2: Explanation 

From the above part, the null hypothesis has not been rejected. Thus, there is a possibility of committing the Type II error.

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