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8.5 Testing a Claim About a Standard Deviation Or Variance

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8.5 Testing a Claim About a Standard Deviation Or Variance 8.5 Testing a Claim about a Standard Deviation or Variance Testing Claims about a Population Standard Deviation or a Population Variance ² Uses the chi-squared distribution from section 7-4 → Requirements: 1. The sample is a simple random sample 2. The population has a normal distribution (n −1)s 2 → Test Statistic for Testing a Claim about or ²: 2 = 2 where n = sample size s = sample standard deviation σ = population standard deviation s2 = sample variance σ2 = population variance → P-values and Critical Values: Use table A-4 with df = n – 1 for the number of degrees of freedom *Remember that table A-4 is based on cumulative areas from the right → Properties of the Chi-Square Distribution: 1. All values of 2 are nonnegative and the distribution is not symmetric 2. There is a different 2 distribution for each number of degrees of freedom 3. The critical values are found in table A-4 (based on cumulative areas from the right) --locate the row corresponding to the appropriate number of degrees of freedom (df = n – 1) --the significance level is used to determine the correct column --Right-tailed test: Because the area to the right of the critical value is 0.05, locate 0.05 at the top of table A-4 --Left-tailed test: With a left-tailed area of 0.05, the area to the right of the critical value is 0.95 so locate 0.95 at the top of table A-4 --Two-tailed test: Divide the significance level of 0.05 between the left and right tails, so the areas to the right of the two critical values are 0.975 and 0.025. Locate 0.975 and 0.025 at the top of table A-4 → Ex #1: The industrial world shares this common goal: Improve quality by reducing variation. Quality control engineers want to ensure that a product has an acceptable mean, but they also want to produce items of consistent quality so that there will be few defects. The Newport Bottling Company had been manufacturing cans of cola with amounts having a standard deviation of 0.051 oz. A new bottling machine is tested, and a simple random sample of 24 cans results in the amounts (in ounces) listed below. (Those 24 amounts have a standard deviation of s = 0.039 oz). Use a 0.05 significance level to test the claim that cans of cola from the new machine(n −1)s 2 have amounts 2 = with a standard deviation that is less than 0.051 oz. 2 11.98 11.98 11.99 11.98 11.90 12.02 11.99 11.93 12.02 12.02 12.02 11.98 12.01 12.00 11.99 11.95 11.95 11.96 11.96 12.02 11.99 12.07 11.93 12.05 --Check requirements: simple random sample, histogram & normal quantile plot show that the sample appears to come from a population with a normal distribution --Critical value method of testing hypotheses: Step 1: The original claim: < 0.051 Step 2: If the original claim is false, then ≥ 0.051 Step 3: H0: = 0.051 H1: < 0.051 Step 4: The significance level is = 0.05 Step 5: Because the claim is made about , we use the chi-square distribution and we have a left-tailed test (24 −1)(0.039) 2 Step 6: The test statistic is = = 13.450 (0.051) 2 Using table A-4, df = 23 and the column corresponding to 0.95, we find the critical value = 13.091. Step 7: Because the test statistic is not in the critical region, we fail to reject the null hypothesis. Step 8: There is not sufficient evidence to support the claim that the standard deviation of amounts with the new machine is less than 0.051 oz. → Ex #2: The Skytek Avionics company uses a new production method to manufacture aircraft altimeters. A simple random sample of new altimeters resulted in the errors listed below. Use a 0.01 significance level to test the claim that the new production method has errors with a standard deviation greater than 32.2 feet, which was the standard deviation for the old production method. If it appears that the standard deviation is greater, does the new production method appear to be better or worse than the old method? Should the company take any action? (n −1)s 2 2 = -42 78 -22 -72 -45 15 17 51 -5 -53 2 -9 -109 --Critical value method of testing hypotheses: Step 1: The original claim: > 32.2 Step 2: If the original claim is false, then ≤ 32.2 Step 3: H0: = 32.2 H1: > 32.2 Step 4: The significance level is = 0.01 Step 5: Because the claim is made about , we use the chi-square distribution and we have a right tailed test (12 −1)(52.441)2 Step 6: The test statistic is = = 29.176 (32.2)2 Using table A-4, df = 11 and the column corresponding to 0.01, we find the critical value = 24.725 Step 7: Because the test statistic is in the critical region, we reject the null hypothesis. Step 8: There is sufficient evidence to support the claim that the new production method has errors with a standard deviation greater than 32.2 feet. The variation appears to be greater than the previous method so the new methods appears to be worse because there will be more altimeters that will have larger errors. The company should take immediate action. HW 8.5: p.428-431 #1, 4, 10, 13 .
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