Question

the measurements of the diameters (in inches) of 12 randomly chosen golf balls are listed. at $\alpha = 0.05$, is there enough evidence to reject the claim that the standard deviation of the measurements of these diameters is 0.005? assume the population is normally distributed. 1.677 1.682 1.681 1.679 1.677 1.681 1.682 1.678 1.677 1.682 1.679 1.678 click the icon to view the chi - square distribution table. (a) write the claim mathematically and identify $h_0$ and $h_a$. choose the correct answer below. \\(\bigcirc\\) a. $h_0: \sigma > 0.005$; $h_a: \sigma \leq 0.005$ (claim) \\(\bigcirc\\) b. $h_0: \sigma \leq 0.005$ (claim); $h_a: \sigma > 0.005$ \\(\bigcirc\\) c. $h_0: \sigma = 0.005$ (claim); $h_a: \sigma \
eq 0.005$ \LXI0 d. $h_0: \sigma \geq 0.005$; $h_a: \sigma < 0.005$ (claim) (b) find the critical value(s). $\chi_0^2 = \square$ (round to three decimal places as needed. use a comma to separate answers as needed.)

Explanation:

Part (a)

The claim is that the standard deviation \( \sigma = 0.005 \). The null hypothesis \( H_0 \) contains the claim (equality), and the alternative hypothesis \( H_a \) is the complement (two - tailed, \( \sigma
eq0.005 \)) since we are testing if there is enough evidence to reject the claim that \( \sigma = 0.005 \). Option A has the claim in the alternative, Option B has a one - tailed alternative with a different claim, Option D also has a wrong claim and a one - tailed alternative.

Part (b)

Step 1: Determine the degrees of freedom

The sample size \( n = 12 \). The degrees of freedom for a chi - square test for standard deviation is \( df=n - 1\). So, \( df=12 - 1=11 \).

Step 2: Determine the significance level and the type of test

The significance level \( \alpha = 0.05 \). Since the test is two - tailed (because \( H_a:\sigma
eq0.005 \)), we split the significance level into two tails. So, \( \alpha/2=0.025 \) and \( 1-\alpha/2 = 0.975 \).

Step 3: Find the critical values from the chi - square distribution table

We need to find \( \chi_{0.975,11}^{2} \) and \( \chi_{0.025,11}^{2} \) from the chi - square distribution table.

• For \( \chi_{0.975,11}^{2} \): Looking at the chi - square distribution table with \( df = 11 \) and \( \alpha/2=0.975 \), we find that \( \chi_{0.975,11}^{2}=3.816 \).
• For \( \chi_{0.025,11}^{2} \): Looking at the chi - square distribution table with \( df = 11 \) and \( \alpha/2 = 0.025 \), we find that \( \chi_{0.025,11}^{2}=21.920 \).

Answer:

C. \( H_0: \sigma = 0.005 \) (Claim); \( H_a: \sigma
eq 0.005 \)