Cuarters are cunently minted with weights having a mean of 5.185 and a standard deviation of 0.082. New equicment is be ig tested in an attempt to improve quality oy reduc ing variation. A simple rancom sample of 20 cuarters is obtained from those manufactured with the new equipment and this sample has a standard deviation of 0.043. Use a 0.05 significance level to teet the claim that quarters manufisctured with the new equiprrent have weights with a standard deviation less than 0062 . Doses the new equipment appear to be effective in reducing the wariation of weights? What can be said about the effectiveness of the new equipment A. Since Hn?:s=0.082 is not rejected, the new equipment is not more effective. B. Since H0?: o 0.082 is rejected the nev equipment is more effective. C. Since H0?: a =0.082 is not rejected. the new equipment is not more effective. D. Since H0?:p=0. DE2 is rejected. the new equipment is more effective. A teacher claims that her students’ test scores are getting more consistent and now have a lower variation than 2.31. the variation in previous terms. She conducts a hypothesis test. She calculates her test statistic to be ?2=13.641 She looks up the critical value for this test and finds it to be x2=20.546. What can she conclude? Hint: Set-up Ha and H1? first, draw a sketch and read the answer choices carefully! A. The test statistic does not fall in the critisal irejectionj region for this tost. Therofore, there is sufficient evidence to support her claim. B. The test statistic fals in the critical frejecticni region for this test. Therefore, there is sufficient evidenee to support her claim. z. The test statistic fals in the eritical [rejectieni) region for this lest. Therefore, there is not sufficient evidenee to support her claim. D. The test statistic does not fall in the critical irejectioni region for this test. Therefore, there is not sumcient evicence to support her claim.
The correct answer and explanation is:
Problem 1: Testing if the new equipment reduces the standard deviation of quarter weights
Given Data:
- Null hypothesis ((H_0)): (\sigma = 0.082)
- Alternative hypothesis ((H_a)): (\sigma < 0.082)
- Sample size ((n)): 20
- Sample standard deviation ((s)): 0.043
- Significance level ((\alpha)): 0.05
Test Statistic:
This is a chi-square test for variance:
[
\chi^2 = \frac{(n – 1)s^2}{\sigma_0^2}
]
[
\chi^2 = \frac{(20 – 1)(0.043^2)}{0.082^2} = \frac{19(0.001849)}{0.006724} = 5.222
]
Degrees of freedom ((df)):
[
df = n – 1 = 20 – 1 = 19
]
Critical Value:
Using a chi-square table at (\alpha = 0.05) for (df = 19) in a left-tailed test:
[
\chi^2_{\text{critical}} = 10.117
]
Decision:
Since (5.222 < 10.117), the test statistic falls in the rejection region, and we reject (H_0).
Conclusion:
The new equipment significantly reduces the standard deviation of weights. Correct answer: B. Since (H_0: \sigma = 0.082) is rejected, the new equipment is more effective.
Problem 2: Testing the teacher’s claim about reduced variation in test scores
Given Data:
- Null hypothesis ((H_0)): (\sigma^2 = 2.31)
- Alternative hypothesis ((H_a)): (\sigma^2 < 2.31)
- Test statistic: (\chi^2 = 13.641)
- Critical value: (\chi^2_{\text{critical}} = 20.546)
Decision:
Since (13.641 < 20.546), the test statistic does not fall in the critical rejection region.
Conclusion:
There is sufficient evidence to support the teacher’s claim that the test scores have less variation. Correct answer: A. The test statistic does not fall in the critical (rejection) region for this test. Therefore, there is sufficient evidence to support her claim.
Explanations Combined
In hypothesis testing, we assess whether a sample provides enough evidence to reject the null hypothesis. Both problems involve chi-square tests for variance.
In Problem 1, the null hypothesis claims that the standard deviation is 0.082, while the alternative claims it is less. The chi-square test compares the sample variance to the hypothesized population variance under the assumption (H_0) is true. A test statistic smaller than the critical value indicates less variance than hypothesized. With a test statistic of 5.222 falling below the critical value of 10.117, we conclude that the new equipment reduces variation, suggesting improved manufacturing consistency.
In Problem 2, the teacher claims reduced variation in test scores compared to a previous term’s variance of 2.31. Her test statistic (13.641) is less than the critical value (20.546), indicating no evidence to reject (H_0). This supports her claim of reduced variation, as the data aligns with the alternative hypothesis.
For both tests, understanding the rejection region, critical value, and the relationship between the test statistic and (H_0) is essential for correct conclusions.