A machine that puts corn flakes into boxes is adjusted to put an average of 15.5 ounces into each box, with standard deviation of 0.23 ounce. If a random sample of 15 boxes gave a sample standard deviation of 0.38 ounce, do these data support the claim that the variance has increased and the machine needs to be brought back into adjustment? (Use a 0.01 level of significance.)(i) Find the sample test statistic. (Round your answer to two decimal places.)(ii) Find or estimate the P-value of the sample test statistic.P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005(iii) Conclude the test.Since the P-value ≥ α, we fail to reject the null hypothesis.Since the P-value < α, we reject the null hypothesis.    Since the P-value < α, we fail to reject the null hypothesis.Since the P-value ≥ α, we reject the null hypothesis.

Question
Asked Dec 6, 2019
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A machine that puts corn flakes into boxes is adjusted to put an average of 15.5 ounces into each box, with standard deviation of 0.23 ounce. If a random sample of 15 boxes gave a sample standard deviation of 0.38 ounce, do these data support the claim that the variance has increased and the machine needs to be brought back into adjustment? (Use a 0.01 level of significance.)

(i) Find the sample test statistic. (Round your answer to two decimal places.)


(ii) Find or estimate the P-value of the sample test statistic.
P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

(iii) Conclude the test.
Since the P-value ≥ α, we fail to reject the null hypothesis.Since the P-value < α, we reject the null hypothesis.    Since the P-value < α, we fail to reject the null hypothesis.Since the P-value ≥ α, we reject the null hypothesis.

 
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Expert Answer

Step 1

(i)

 

The claim of the problem is the variance is increased.

 

From the given information, the population mean is 15.5 ounces and the population standard deviation is 0.23 ounces. There are 15 boxes with sample standard deviation 0.38.

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Null Hypothesis: H,:o² =0.0529. Alternative Hypothesis: H:o² > 0.0529.

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Step 2

The test statistic is 38.22 and it is calculated below:

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(n-1)s² (15 – 1)×(0.38)* (0.23) = 38.22

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Step 3

(ii)

 

The degrees of freedom is, n-1=15-1=14.

 

The P-value is 0.0005 and it is obtained by Excel function “=CHISQ.DIST.RT(38.22,14)&rd...

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Hypothesis Testing

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