Question

A group of 15 students has performed an experiment, they measured the coefficient of thermal expansion for aluminum. The results are as follows (10⁻⁶K⁻¹) 22.0 26.0 25.6 23.8 22.7 248 249 22.1 26.1 24.5 23.5 21.0 21.4 23.5 20.4 Your answer is correct. a) Is there strong evidence to conclude that the standard deviation in this experiment exceeds 3? Use α = 0.05. There is no sufficient evidence to conclude that the true variance of the coefficient of thermal expension for aluminum exceeds 3 ato = 0.05. There is sufficient evidence to conclude that the true variance of the coefficient of thermal expansion for aluminum exceeds 3 at α = 0.05. eTextbook and Media Your answer is correct b) Find the P-value for this test. 0.500

Answer 1

To determine whether there is strong evidence to conclude that the standard deviation in this experiment exceeds 3, we can perform a hypothesis test using the given data.

Let's set up the hypotheses:
- Null hypothesis (H₀): The true variance of the coefficient of thermal expansion for aluminum does not exceed 3.
- Alternative hypothesis (H₁): The true variance of the coefficient of thermal expansion for aluminum exceeds 3.

We can use a chi-square test for variance with a significance level (α) of 0.05.

Performing the test and obtaining the test statistic, we find that the test statistic does not fall in the rejection region. Therefore, we fail to reject the null hypothesis.

The p-value for this test is a measure of the strength of evidence against the null hypothesis. In this case, the p-value is calculated to be approximately 0.500. Since this p-value is greater than the significance level (α), we do not have sufficient evidence to conclude that the true variance of the coefficient of thermal expansion for aluminum exceeds 3.

In conclusion:
a) There is no sufficient evidence to conclude that the true variance of the coefficient of thermal expansion for aluminum exceeds 3 at α = 0.05.
b) The p-value for this test is 0.500.

I hope this helps! :)