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Sagot :
Answer:
The pvalue of the test is 0.0083 < 0.1, which means that there is enough evidence to conclude that residents of Legacy Ranch use less water on average
Step-by-step explanation:
The American Water Works Association reports that the per capita water use in a single-family home is 69 gallons per day. Test that the residents use less water.
At the null hypothesis, we test that the mean is 69, that is:
[tex]H_0: \mu = 69[/tex]
At the alternate hypothesis, we test that the mean is less than 69, that is:
[tex]H_a: \mu < 69[/tex]
The test statistic is:
As we have the standard deviation for the sample, we use the t-statistic.
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.
69 is tested at the null hypothesis:
This means that [tex]\mu = 69[/tex]
Thirty-six owners responded, and the sample mean water use per day was 64 gallons with a standard deviation of 8.8 gallons per day.
This means that [tex]n = 36, X = 64, s = 8.8[/tex]
Value of the test statistic:
[tex]t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{64 - 69}{\frac{8.8}{\sqrt{36}}}[/tex]
[tex]t = -3.41[/tex]
Pvalue of the test and decision:
The pvalue of the test is the probability of finding a mean less than 64, which is the pvalue of t = -3.41, found on the t-table with 36 - 1 = 35 degrees of freedom on a one-tailed test.
With calculator help, the pvalue is 0.0083.
The pvalue of the test is 0.0083 < 0.1, which means that there is enough evidence to conclude that residents of Legacy Ranch use less water on average
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