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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 0.5a gallon per flush when compared to its current model. To test this claim, the company randomly selects 30toilets 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 is1.64a gal with a standard deviation of 0.29gal. For the new toilets, the mean amount of water used is 1.09galwith a standard deviation of 0.18gal.

(a) Carry out an appropriate significance test. What conclusion would you draw? (Note that the null hypothesis is notH0:μ1-μ2=0-

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

Short Answer

Expert verified

a)Yes, the data set is providing sufficient evidence.

b) Type ll error.

Step by step solution

01

Part (a) Step 1: Given Information

x¯1=1.64,x¯2=1.09

s1=0.29,s2=0.18

n1=30,n2=30

02

Part (a) Step 2: Explanation

Test statistic formula is:

t=x¯1-x¯2x17n1+x22n2

The null and alternative hypotheses for the provided case are:

H0:μ1-μ2=0.5

H1:μ1-μ2>0.5

The test statistic is computed as:

t=x¯1-x¯2-μ1-μ2s12n1+s22n2

=1.64-1.09-(0.5)0.29230+0.18230

=0.802

The degree of freedom is calculated as:

df=minn1-1,n2-1=min(30-1,30-1)=29

The p-value:

P-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\%.

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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