Question

A petroleum company fills large drums of oil to distribute to customers. In a random sample of 35 drums, the mean number of gallons of oil is 202.52 and the standard deviation is 1.23. The company would like to ensure its customers that the average number of gallons of oil in the drums that the company distributes is greater than 200 gallons. We will conduct a statistical test to evaluate this claim.

1. Formulate the null hypothesis (in words) to evaluate this claim.

2. Formulate the alternative hypothesis (in words) to evaluate this claim.

3. What is μ0 in the statistical test?

4. What is the population parameter that the statistical test is concerned with?

5. Which sample statistic can be used to estimate the population parameter that the statistical tests is testing?

6. Is using the parametric tests discussed in class applicable? Explain why or why not.

7. What values of x̄ would lead you to reject the null hypothesis at a confidence level of 90%?

   Calculate this based on the t-test and z-test.

8. What values of x̄ would lead you to reject the null hypothesis at a confidence level of 95%?

   Calculate this based on the t-test and z-test.

9. What values of x̄ would lead you to reject the null hypothesis at a confidence level of 99%?

   Calculate this based on the t-test and z-test.

   x̄−μ0

10. Suppose we transform x̄ into the test statistic T = s/√n . What values of T would lead you

    to reject the null hypothesis at a confidence level of 90%? Calculate this based on the t-test and z-test.

    x̄−μ0

11. Suppose we transform x̄ into the test statistic T = s/√n . What values of T would lead you

    to reject the null hypothesis at a confidence level of 95%? Calculate this based on the t-test and z-test.

    x̄−μ0

12. Suppose we transform x̄ into the test statistic T = s/√n . What values of T would lead you

    to reject the null hypothesis at a confidence level of 99%? Calculate this based on the t-test and z-test.

13. What is the p-value of this test? Calculate it based on x̄ and also based on the transformed statistic, using both the t and normal distributions.

14. What is the minimum confidence level for which you would reject the null hypothesis? Answer this using the p-values for both the t and normal distribution.

Answer 1

Solution:-

1)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

a)

Null hypothesis: The average number of gallons of oil in the drums that the company distributes is 200 gallons.

Null hypothesis: u < 200

b)

Alternative hypothesis: The average number of gallons of oil in the drums that the company distributes is greater than 200 gallons.
Alternative hypothesis: u > 200

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

c) The μ₀ in the statistical test is 200.

d) The population parameter that the statistical test is concerned with is  the population average number of gallons of oil in the drums that the company distributes.

e) The sample statistic can be used to estimate the population parameter that the statistical tests is testing is sample mean of 35 drums, the sample mean number of gallons of oil is 202.52.

f) The values of x̄ would lead you to reject the null hypothesis at a confidence level of 90% is

D.F = 34

t_critical = 1.307

t = (x - u) / SE

x̄ = 200.272

The values of x̄ would lead you to reject the null hypothesis at a confidence level of 95% is

D.F = 34

t_critical = 1.691

t = (x - u) / SE

x̄ = 200.352

The values of x̄ would lead you to reject the null hypothesis at a confidence level of 99% is

D.F = 34

t_critical = 2.441

t = (x - u) / SE

x̄ = 200.508

SE = s / sqrt(n)

S.E = 0.20791
DF = n - 1

D.F = 34

t = (x - u) / SE

t = 12.12

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of 12.12.

Thus the P-value in this analysis is less than 0.0001.

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.