Questions: 2. The amount of water consumed each week by Montana residences is normally distributed. A simple random sample of 10 residences was taken with a sample mean of 120.3 gallons and a standard deviation of 10.0 gallons. Test the claim at the 0.10 SL that the average amount of water consumed is not 125 gallons.

Test the claim that the average amount of water consumed is not 125 gallons.

Define the null and alternative hypotheses.

Null hypothesis (\(H_0\)): \(\mu = 125\) gallons
Alternative hypothesis (\(H_1\)): \(\mu \neq 125\) gallons

Determine the test statistic.

Since the population standard deviation is unknown and the sample size is small (\(n = 10\)), use a t-test.
The t-statistic formula is:
\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]

Calculate the test statistic.

Given:
\(\bar{x} = 120.3\), \(\mu_0 = 125\), \(s = 10.0\), \(n = 10\)
\[ t = \frac{120.3 - 125}{10.0 / \sqrt{10}} \approx -1.486 \]

Determine the critical value(s).

Degrees of freedom: \(df = n - 1 = 9\)
Significance level: \(\alpha = 0.10\)
For a two-tailed test, critical t-values for \(\alpha/2 = 0.05\) and \(df = 9\) are approximately \(\pm 1.833\).

Make a decision.

Since \(-1.833 < -1.486 < 1.833\), the test statistic falls within the non-rejection region. Therefore, we fail to reject the null hypothesis.

State the conclusion.

At the 0.10 significance level, there is not enough evidence to support the claim that the average amount of water consumed is not 125 gallons.

\(\boxed{\text{Fail to reject } H_0: \text{ not enough evidence to claim the average is not 125 gallons.}}\)

\(\boxed{\text{Fail to reject } H_0: \text{ not enough evidence to claim the average is not 125 gallons.}}\)