Question

The US Department of Energy reported that 45% of homes were heated by natural gas. A random sample of 315 homes in Oregon found that 169 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 answers to at least 4 decimal places.

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

H0:

HA:

Based on the hypotheses, find the following:
Test Statistic =
p-value =

Answer 1

The correct hypotheses for the test are H0: p = 0.45 and HA: p ≠ 0.45. The test statistic is 2.3708 and the p-value is approximately 0.0177.

Step-by-step explanation:

To test the claim that the proportion of homes in Oregon that were heated by natural gas is different than what was reported, we need to set up the hypotheses:

H0: p = 0.45 (The proportion of homes in Oregon heated by natural gas is the same as reported)

HA: p ≠ 0.45 (The proportion of homes in Oregon heated by natural gas is different than reported)

Next, we need to find the test statistic and p-value:

The test statistic is calculated using the formula: z = (p - P) / sqrt(P(1 - P) / n),

where p is the sample proportion, P is the proportion reported by the US Department of Energy, and n is the sample size.

Using the given values, we have: p = 169/315 = 0.537, P = 0.45, and n = 315.

Substituting these values into the formula, we get: z = (0.537 - 0.45) / sqrt(0.45(1 - 0.45) / 315) = 2.3708

The p-value is the probability of observing a test statistic as extreme or more extreme than the one calculated, assuming the null hypothesis is true. To find the p-value, we can use a standard normal distribution table or a calculator. In this case, the p-value is approximately 0.0177.