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Question
the us department of energy reported that 49% of homes were heated by natural gas. a random sample of 316 homes in oregon found that 158 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 10% 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.)
( h_{0} ): select an answer ?
( h_{1} ): select an answer ?
based on the hypotheses, compute the following:
b. test statistic =
c. p - value =
d. based on the above we choose to select an answer
e. the correct summary would be: select an answer that the proportion of homes in oregon that were heated by natural gas is different than what the doe reported value of 49%.
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Step1: State the hypotheses
The null hypothesis \(H_0\) is that the proportion \(p\) is equal to the reported proportion. The alternative hypothesis \(H_1\) is that the proportion \(p\) is different from the reported proportion.
\(H_0: p = 0.49\)
\(H_1: p
eq0.49\)
Step2: Calculate the sample proportion \(\hat{p}\)
The sample proportion \(\hat{p}=\frac{x}{n}\), where \(x = 158\) and \(n=316\).
\(\hat{p}=\frac{158}{316}=0.5\)
Step3: Calculate the test statistic \(z\)
The formula for the test statistic in a proportion test is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\)
Substitute \(p = 0.49\), \(\hat{p}=0.5\), and \(n = 316\) into the formula:
Step4: Calculate the p - value
Since this is a two - tailed test (\(H_1: p
eq0.49\)), the p - value is \(2\times(1 - P(Z<|z|))\)
Using a standard normal table or calculator, \(P(Z < 0.3556)\approx0.6381\)
\(p - value=2\times(1 - 0.6381)=2\times0.3619 = 0.7238\)
Step5: Make a decision
Since the p - value (\(0.7238\)) is greater than the significance level \(\alpha=0.10\), we fail to reject the null hypothesis.
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a. \(H_0: p = 0.49\); \(H_1: p
eq0.49\)
b. \(0.3556\)
c. \(0.7238\)
d. Fail to reject \(H_0\)
e. There is not sufficient evidence