Question

The American Water Works Association reports that the per capita water use in a single-family home is 67 gallons per day. Legacy Ranch is a relatively new housing development. The builders installed more efficient water fixtures, such as low-flush toilets, and subsequently conducted a survey of the residences. Twenty-three owners responded, and the sample mean water use per day was 63 gallons with a standard deviation of 8.1 gallons per day. At the 0.025 level of significance, is that enough evidence to conclude that residents of Legacy Ranch use less water on average?a. What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.) Reject H0: µ ≥ 67 when the test statistic is (Click to select)less thanmore than .b. The value of the test statistic is . (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)c. What is your decision regarding H0?

Answer 1

Answer:

We conclude that the null hypothesis is rejected which means residents of Legacy Ranch use less water on average.

Step-by-step explanation:

We are given that the American Water Works Association reports that the per capita water use in a single-family home is 67 gallons per day.

Twenty-three owners responded, and the sample mean water use per day was 63 gallons with a standard deviation of 8.1 gallons per day.

Let $\mu$ = average water usage by residents of Legacy Ranch.

So, Null Hypothesis, $H_0$ : $\mu \geq$ 67 gallons     {means that the residents of Legacy Ranch uses more or equal to 67 gallons water on average}

Alternate Hypothesis, $H_A$ : $\mu$ < 67 gallons    {means that the residents of Legacy Ranch use less water on average}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

                                 T.S.  = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample mean water use per day = 63 gallons

             s = sample standard deviation = 8.1 gallons

             n = sample of responded = 23

So, test statistics  =  $\frac{63-67}{\frac{8.1}{\sqrt{23} } }$  ~ $t_2_2$

                               =  -2.368

The value of the test statistics is -2.368.

Now at 0.025 significance level, the t table gives critical value of -2.074 at 22 degree of freedom for left-tailed test. Since our test statistics is less than the critical values of t as -2.368 < -2.074, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the residents of Legacy Ranch use less water on average.