Answer to Question #239563 in Statistics and Probability for huzafi

Question

Expensive test borings were made in an oil shale area to determine if the mean yield of oil per ton of shale rock is greater than 4.5 barrels. To check it a random sample of 5 borings made with mean 4.9 and standard deviation 0.64 barrels. Perform test of hypothesis at 5% level of significance?

Answer 1

Null and alternative hypotheses :

$H_0: \mu ≤4.5 \\ H_1: \mu >4.5$

Test statistic :

To test the hypothesis the most appropriate test is one sample t-test. The test statistic is given as follows:

$t = \frac{\bar{x} - \mu}{s/ \sqrt{n}}$

Where, $\bar{x}$ is sample mean, s is sample standard deviation, n is sample size and μ is hypothesized value of population mean under H₀.

We are given the following values :

$\bar{x}= 4.9 \\ s = 0.64 \\ n = 5 \\ μ = 4.5 \\ t =\frac{4.9-4.5}{0.64/ \sqrt{5}} = 1.3975$

The value of the test statistic is 1.3975.

P-value :

Our test is right-tailed test. The p-value for right-tailed t-test is given as follows :

P-value = P(T > t)

P-value = P(T > 1.3975)

P-value = 0.1174

Decision :

Significance level = 5% = 0.05

P-value = 0.1174

(0.1174 > 0.05)

Since, p-value is greater than the significance level of 5%, therefore we shall be fail to reject the null hypothesis (H₀) at 5% significance level.

Conclusion :

At 5% significance level, there is not sufficient evidence to conclude that mean yield of oil per ton of shale rock is greater than 4.5 barrels.