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. 
<u><em>Let </em></u> <u><em> = average water usage by residents of Legacy Ranch.</em></u>
<u><em> = average water usage by residents of Legacy Ranch.</em></u>
So, Null Hypothesis,  :
 :  67 gallons     {means that the residents of Legacy Ranch uses more or equal to 67 gallons water on average}
 67 gallons     {means that the residents of Legacy Ranch uses more or equal to 67 gallons water on average}
Alternate Hypothesis,  :
 :  < 67 gallons    {means that the residents of Legacy Ranch use less water on average}
 < 67 gallons    {means that the residents of Legacy Ranch use less water on average}
The test statistics that will be used here is <u>One-sample t test statistics</u> as we don't know about population standard deviation;
                                  T.S.  =  ~
  ~ 
where,  = sample mean water use per day = 63 gallons
 = sample mean water use per day = 63 gallons
              s = sample standard deviation = 8.1 gallons
              n = sample of responded = 23
So, <u><em>test statistics</em></u>  =   ~
  ~ 
                                =  -2.368
The value of the test statistics is -2.368.
<em>Now at 0.025 significance level, the </em><u><em>t table gives critical value of -2.074 at 22 degree of freedom for left-tailed test</em></u><em>. 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 </em><u><em>which we reject our null hypothesis</em></u><em>.</em>
Therefore, we conclude that the residents of Legacy Ranch use less water on average.