Question

The American Water Works Association reports that the per capita water use in a single-family home is 63 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-eight owners responded, and the sample mean water use per day was 60 gallons with a standard deviation of 8.9 gallons per day. At the 0.01 level of significance, is that enough evidence to conclude that residents of Legacy Ranch use less water on average

Answer 1

Answer:

$t=\frac{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784$    

$df=n-1=28-1=27$  

$p_v =P(t_{(27)}$  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is less than 63 gallons per day at 1% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=60$ represent the sample mean

$s=8.9$ represent the sample standard deviation

$n=28$ sample size  

$\mu_o =68$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is less than 63, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 63$  

Alternative hypothesis:$\mu < 63$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$t=\frac{60-63}{\frac{8.9}{\sqrt{28}}}=-1.784$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=28-1=27$  

Since is a one side lower test the p value would be:  

$p_v =P(t_{(27)}$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is less than 63 gallons per day at 1% of signficance.