Step 1 :Define the given values: number of homes in the sample (n) = 324, number of homes heated by natural gas (x) = 126, and the reported national proportion (p0) = 0.4.
Step 2 :Calculate the sample proportion (p_hat) = x/n = 126/324 = 0.3888888888888889.
Step 3 :Calculate the standard error (se) = sqrt[p0*(1-p0)/n] = sqrt[0.4*(1-0.4)/324] = 0.02721655269759087.
Step 4 :Calculate the test statistic (z) = (p_hat - p0) / se = (0.3888888888888889 - 0.4) / 0.02721655269759087 = -0.4082482904638636.
Step 5 :Calculate the p-value using the standard normal distribution. The p-value is the probability of observing a test statistic as extreme as the one calculated, assuming the null hypothesis is true. The p-value is approximately 0.6830913983096083.
Step 6 :Interpret the results. The test statistic is approximately \(-0.4082\) and the p-value is approximately \(0.6831\). This means that if the null hypothesis is true (i.e., the proportion of homes heated by natural gas in Oregon is the same as the reported national proportion), we would expect to see a sample proportion as extreme as the one we observed about 68.31% of the time. This is a relatively large p-value, which suggests that the observed sample proportion is not significantly different from the hypothesized proportion. Therefore, we would not reject the null hypothesis.
Step 7 :Final Answer: The test statistic is approximately \(\boxed{-0.4082}\) and the p-value is approximately \(\boxed{0.6831}\). Based on these results, we would not reject the null hypothesis. The proportion of homes in Oregon that were heated by natural gas is not significantly different from the reported national proportion of 40%.