Question

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.

Short answer

Part a. There is no convincing evidence that the new pressure assisted toilet reduces the average amount of water used by more than $0.5$gallons per flush compare to its current model.

Part b. Type II error.

Step-by-step solution

Part a. Step 1. Given information

$$\begin{gathered} {\overline{x}}_1=1.64 \\ {\overline{x}}_2=1.09 \\ {n}_1=30 \\ {n}_2=30 \\ {s}_1=0.29 \\ {s}_2=0.18 \\ \alpha =0.05 \end{gathered}$$

Part a. Step 2. Explanation

The appropriate hypotheses for this is:

$$\begin{gathered} H_0:{\mu }_1-{\mu }_2=0.5 \\ {H}_a:{\mu }_1-{\mu }_2>0.5 \end{gathered}$$

Now, find the test statistics:

$$\begin{gathered} t=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{{\displaystyle \frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}} \\ =\frac{1.64-1.09-(0.5)}{\sqrt{{\displaystyle \frac{0.{29}^2}{30}}+{\displaystyle \frac{0.{18}^2}{30}}}} \\ =0.802 \end{gathered}$$

Now, the degree of freedom will be:

$$\begin{gathered} df=min(n_1-1,n_2-1) \\ =min(30-1,30-1) \\ =29 \end{gathered}$$

So the P-value will be:

$0.20<P<0.25$

And we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, then,

$P>0.05\Rightarrow FailtoReject{H}_0$

Thus, we conclude that there is no convincing evidence that the new pressure assisted toilet reduces the average amount of water used by more than $0.5$gallons per flush compare to its current model.

Part b. Step 1. Explanation

We conclude in part (b) that,

There is no convincing evidence that the new pressure assisted toilet reduces the average amount of water used by more than gallons per flush compare to its current model.

A type I error occurs if we reject a null hypothesis when the null hypothesis is true. And the Type II error occurs if we fails to reject the null hypothesis when the null hypothesis is false.

Thus, in this case we fail to reject the null hypothesis then it is a Type II error.