At a shooting range, instructors can determine if a shooter is consistently missing the target because of the gun sight or because of the shooter’s
ability. If a gun’s sight is off, the variance of the distances between the shots and the center of the shot pattern will be small (even if the shots are not
in the center of the target). A student claims that it is the sight that is off, not his aim, and wants the instructor to confirm his claim. If a skilled shooter
fires a gun at a target multiple times, the distances between the shots and the center of the shot pattern, measured in centimeters (cm), will have a
variance of less than 0.33 . After the student shoots 30 shots at the target, the instructor calculates that the distances between his shots and the
center of the shot pattern, measured in cm, have a variance of 0.21 . Does this evidence support the student’s claim that the gun’s sight is off? Use a
0.10 level of significance. Assume that the distances between the shots and the center of the shot pattern are normally distributed.
Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places.

Question

UpStudy AI · Accepted Answer

To test the student's claim that the variance of the distances between the shots and the center of the shot pattern is less than 0.33, we can use a chi-squared test for variance.

The null hypothesis ($$H_0$$) and the alternative hypothesis ($$H_a$$) are defined as follows:

- $$H_0: \sigma^2 \geq 0.33$$ (the variance is greater than or equal to 0.33)
- $$H_a: \sigma^2 < 0.33$$ (the variance is less than 0.33)

Given:
- Sample variance ($$s^2$$) = 0.21
- Population variance under the null hypothesis ($$\sigma_0^2$$) = 0.33
- Sample size ($$n$$) = 30

The test statistic for the chi-squared test for variance is calculated using the formula:

$$
\chi^2 = \frac{(n - 1) s^2}{\sigma_0^2}
$$

Substituting the values into the formula:

$$
\chi^2 = \frac{(30 - 1) \cdot 0.21}{0.33}
$$

Calculating the numerator:

$$
(30 - 1) \cdot 0.21 = 29 \cdot 0.21 = 6.09
$$

Now, substituting this into the chi-squared formula:

$$
\chi^2 = \frac{6.09}{0.33} \approx 18.45454545
$$

Now, rounding to three decimal places:

$$
\chi^2 \approx 18.455
$$

Thus, the value of the test statistic is approximately **18.455**.
The value of the test statistic is approximately 18.455.

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