Answered

The US Department of Energy reported that $49 \%$ of homes were heated by natural gas. A random sample of 308 homes in Oregon found that 165 were heated by natural gas. Test the claim that the proportion of homes in Oregon that were heated by natural gas is different than what was reported. Use a 1\% significance level. Give answers to at least 4 decimal places.


AWhat are the correct hypotheses? (Select the correct symbols and use decimal values not percentages.)

$H_0$: □ p □ 0.49

$H_1$: □ p □ 0.49



Based on the hypotheses, compute the following:

BTest Statistic = □
Cp-value = □
DBased on the above, we choose to □ [Reject the null hypothesis / Fail to reject the null hypothesis]
EThe correct summary would be: □ that the proportion of homes in Oregon that were heated by natural gas is different than the DOE reported value of $49 \%$.

Verified Answer :

  • Hypotheses: \(H_0: p = 0.49\), \(H_1: p eq 0.49\).
  • Test Statistic: \(z \approx \boxed{1.6049}\).
  • p-value: \(p \approx \boxed{0.1085}\).
  • Decision: Fail to reject \(H_0\).
  • Summary: There is not sufficient evidence to conclude that the proportion is different.

Explanation

  1. State the Hypotheses

First, we define our hypotheses. The null hypothesis, \(H_0\), represents the status quo or the claim being tested against, which is that the proportion of homes heated by natural gas in Oregon is the same as the reported national figure of \(0.49\). The alternative hypothesis, \(H_1\), is that the proportion is different, indicating a two-tailed test.

  • \(H_0: p = 0.49\)
  • \(H_1: p eq 0.49\)
  1. Calculate the Test Statistic

To determine if our sample data significantly deviates from the reported \(49\%\), we calculate the test statistic \(z\).

  • Sample size \(n = 308\)
  • Observed successes \(x = 165\)
  • Sample proportion \(\hat{p} = \frac{165}{308} \approx 0.5357\)
  • Population proportion under \(H_0\), \(p_0 = 0.49\)
  • Standard Error \(SE = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.49 \times 0.51}{308}} \approx 0.0285\)
  • Test Statistic \(z = \frac{\hat{p} - p_0}{SE} = \frac{0.5357 - 0.49}{0.0285} \approx 1.6049\)
  1. Calculate the p-value

The p-value tells us the probability of observing a sample proportion as extreme as ours (or more so), assuming the null hypothesis is true. For a two-tailed test with a \(z\)-score of approximately \(1.6049\):

  • \(p\text{-value} = 2 \times P(Z > |1.6049|) \approx 2 \times 0.0543 = 0.1085\)
  1. Decision and Conclusion

We compare the calculated p-value of \(0.1085\) with our significance level \(\alpha = 0.01\). Since \(0.1085 > 0.01\), we fail to reject the null hypothesis. There is not enough statistical evidence at the \(1\%\) significance level to conclude that the proportion of homes in Oregon heated by natural gas is different from the \(49\%\) reported by the DOE.

Examples

This statistical hypothesis testing process is vital for quality control and auditing in government reporting. For example, if an agency claims that a policy impacts 49% of the population, researchers use this method to verify if local data from a specific state significantly deviates from that claim, helping to ensure transparency and accuracy in public data.

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