Question

The American Water Works Association reports that the per capita water use in a single-family home is 77 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 73 gallons with a standard deviation of 7.6 gallons per day.

At the 0.10 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 ≥ 79 and fail to reject H1: μ < 79 when the test statistic is less than what? .
(b) The value of the test statistic is . (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

Answer 1

Explanation:

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 77

For the alternative hypothesis,

H1: µ < 77

This is a left tailed test

Therefore, we would find the critical value corresponding to 1 - α and reject the null hypothesis if the test statistic is less than the negative of the table value.

1 - α = 1 - 0.1 = 0.9

The negative critical value is - 1.314

Reject H0: μ = 77 and fail to reject H1: μ < 77 when the test statistic is less than - 1.314

b) Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 28

Degrees of freedom, df = n - 1 = 28 - 1 = 27

t = (x - µ)/(s/√n)

Where

x = sample mean = 73

µ = population mean = 77

s = samples standard deviation = 7.6

t = (73 - 77)/(7.6/√28) = - 2.79

Since - 2.79 is lesser than - 1.314, then we would reject the null hypothesis.