91影视

$$\begin{gathered} {\stackrel{炉}{x}}_1=1.64 \\ {\stackrel{炉}{x}}_2=1.09 \\ {n}_1=30 \\ {n}_2=30 \\ {s}_1=0.29 \\ {s}_2=0.18 \\ 伪=0.05 \end{gathered}$$

The appropriate hypotheses for this is:

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

Now, find the test statistics:

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

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,

$P>0.05鈬�FailtoReject{H}_0$

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.

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.