Question

The average gasoline price of one of the major oil companies has been $3.00 per gallon. Because of shortages in production of crude oil, it is believed that there has been a significant increase in the average price. In order to test this belief, we randomly selected a sample of 36 of the company’s gas stations and determined that the average price for the stations in the sample was $3.06. Assume that the standard deviation of the population is $0.09. Exam Review 3 2

(a) State the null and the alternative hypothesis.
(b) Test the claim at α = 0.05.
(c) What is the p-value associated with the above sample results

Answer 1

The null and alternative hypotheses are stated, the claim is tested using a one-sample t-test, and the p-value is discussed.

Step-by-step explanation:

(a) Null and alternative hypothesis:

The null hypothesis (H0) states that there has been no significant increase in the average price of gasoline. The alternative hypothesis (Ha) states that there has been a significant increase in the average price of gasoline.

(b) Testing the claim at α = 0.05:

To test the claim, we can use a one-sample t-test since we have the sample mean, population mean, sample size, and population standard deviation. We can calculate the test statistic (t-value) using the formula:

t = (sample mean - population mean) / (population standard deviation / sqrt(sample size))

If the absolute value of the t-value is greater than the critical value from the t-table at α = 0.05 and (sample mean - population mean) > 0, we reject the null hypothesis and conclude that there has been a significant increase in the average price of gasoline.

(c) The p-value:

The p-value is the probability of obtaining a sample mean at least as extreme as the observed one, assuming the null hypothesis is true. To find the p-value, we can use the t-distribution table or calculator. If the p-value is less than α = 0.05, we reject the null hypothesis.

Answer 2

(a)
`H_(0): \mu = 3.00` vs
`H_(1): \mu > 3.00` (upper-tail alternative)

(b) We reject the null hypothesis at the significance level of 0.05.

(c) The p-value is 3.167124e-05

Step-by-step explanation:

`H_(0): \mu = 3.00` vs
`H_(1): \mu > 3.00` (upper-tail alternative)

We have
`\bar{x} = 3.06`,
`\sigma = 0.09` and n = 36. We have a large sample and our test statistic is

`Z = \frac{\bar{X}-3.00}{\sigma/√(n)}` which shoud be normal standard when
`H_(0)` is true. We have observed

`z = (3.06-3.00)/(0.09/√(36)) = 4`.

We should use the significance level
`\alpha = 0.05`. The 95th quantile of the standard normal distribution is
`z_(0.95) = 1.6449` and the rejection region is given by RR = z . Because the observed value 4 is greter than 1.6449, we reject the null hypothesis at the significance level of 0.05.

The p-value is computed as P(Z > 4) = 3.167124e-05