Question

The US Department of Energy reported that 45% of homes were heated by natural gas. A random sample of 300 homes in Oregon found that 170 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.

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

H₀: Select an answer s s² p̂ σ² σ p μ x̄  Select an answer > ≤ < = ≥ ≠   

H₁: Select an answer x̄ s σ² p̂ s² p σ μ  Select an answer ≤ > = < ≥ ≠   

Based on the hypotheses, find the following:

Test Statistic =  

p-value =

Based on the above we choose to Select an answer Fail to reject the null hypothesis Accept the alternative hypotheis Accept the null hypothesis Reject the null hypothesis

The correct summary would be: Select an answer There is enough evidence to support the claim There is enough evidence to reject the claim There is not enough evidence to reject the claim There is not enough evidence to support the claim  that the proportion of homes in Oregon that were heated by natural gas is different than what the DOE reported value of 45%.

Answer 1

a)

For this study, we should use Select an answer z-test for a population proportion
b)

Below are the null and alternative Hypothesis,
Null Hypothesis: p = 0.45
Alternative Hypothesis: p ≠ 0.45

c)

P = 170/300 = 0.5667

Test statistic,
z= (P -p)/sqrt(p*(1-p)n)
= ( 0.5667 - 0.45)/sqrt(0.45*(1-0.45)/300)
= 4.0630

d)

P-value Approach
P-value = 0.0001

e)
As P-value < = alpha,

f)

reject null hypothesis.

g)

There is enough evidence to support the claim that the proportion of homes in Oregon that were heated by natural gas is different than what the DOE reported value of 45%..