Questions: The US Department of Energy reported that 47% of homes were heated by natural gas. A random sample of 333 homes in Oregon found that 144 were heated by natural gas. Test the claim that proportion of homes in Oregon that were heated by natural gas is different than what was reported. Use a 5% significance level. Give answer to at least 4 decimal places. a. What are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.) b. Test Statistic = c. p-value = d. Based on the above we choose to e. The correct summary would be: that the proportion of homes in Oregon that were heated by natural gas is different than what the DOE reported value of 47%.

Solution Steps

We are testing the claim that the proportion of homes in Oregon heated by natural gas is different from the reported value of \(47\%\). The hypotheses are defined as follows:

• Null Hypothesis (\(H_0\)): \(p = 0.47\)
• Alternative Hypothesis (\(H_1\)): \(p \neq 0.47\)

The sample proportion (\(\hat{p}\)) is calculated as follows:

\[ \hat{p} = \frac{x}{n} = \frac{144}{333} \approx 0.4324 \]

The test statistic \(Z\) is calculated using the formula:

\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \]

Substituting the values:

\[ Z = \frac{0.4324 - 0.47}{\sqrt{\frac{0.47(1 - 0.47)}{333}}} \approx -1.3736 \]

The p-value associated with the test statistic \(Z = -1.3736\) is calculated to be:

\[ \text{P-value} \approx 0.1696 \]

For a significance level of \(\alpha = 0.05\) in a two-tailed test, the critical region is defined as:

\[ Z < -1.96 \quad \text{or} \quad Z > 1.96 \]

Since the calculated p-value \(0.1696\) is greater than the significance level \(\alpha = 0.05\), we fail to reject the null hypothesis.

Final Answer