Chapter+9

03 levelofsignificancetohelpdetermine

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Unformatted text preview: the value of the test statistic z. 2. If z is in the upper tail (z > 0), find the area under the standard normal curve to the right of z. If z is in the lower tail (z < 0), find the area under the standard normal curve to the left of z. 3. Double the tail area obtained in step 2 to obtain the p –value. The rejection rule: Reject H0 if the p­value < α . Critical Value Approach to Critical Value Approach to Two­Tailed Hypothesis Testing The critical values will occur in both the lower and upper tails of the standard normal curve. Use the standard normal probability distribution table to find zα/2 (the z­value with an area of α /2 in the upper tail of the distribution). s The rejection rule is: Reject H0 if z < ­zα/2 or z > zα/2. Example: Glow Toothpaste • Two­Tailed Test About a Population Mean: σ Known The production line for Glow toothpaste is designed to fill tubes with a mean weight of 6 oz. Periodically, a sample of 30 tubes oz. will be selected in order to check the filling process. Glow Quality assurance procedures call for the continuation of the filling process if the sample results are consistent with the assumption that the mean filling weight for the population of toothpaste tubes is 6 oz.; otherwise the process will be adjusted. Example: Glow Toothpaste Example: Glow Toothpaste s Two­Tailed Test About a Population Mean: σ Known Assume that a sample of 30 toothpaste tubes provides a sample mean of 6.1 oz. The population standard deviation is believed to be 0.2 oz. Perform a hypothesis test, at the .03 level of significance, to help determine whether the filling process should continue operating or be stopped and corrected. oz. Glow Two­Tailed Tests About a Population Mean: σ Known p –Value and Critical Value Approaches 1. Determine the hypotheses. 1. Determine the hypotheses. H0: µ = 6 Ha: µ ≠ 6 2. Specify the level of significance. α = .03 3. Compute the value of the test statistic. x − µ0 6.1 − 6 z= = = 2.74 σ / n .2 / 30 Glow Two­Tailed Tests About a Population Mean: σ Known p –Value Approach 4. Compute the p –value. For z = 2.74, cumulative probability = .9969 p–value = 2(1 − .9969) = .0062 5. Determine whether to reject H0. Because p–value = .0062 < α = .03, we reject H0. There is sufficient statistical evidence to infer that the alternative hypothesis is true (i.e. the mean filling weight is not 6 ounces). Glow Two­Tailed Tests About a Population Mean: σ Known Glow p­Value Approach 1/2 p ­value = .0031 1/2 p ­value = .0031 α /2 = .015 α /2 = .015 z = ­2.74 z ­zα/2 = ­2.17 0 zα/2 = 2.17 z = 2.74 Two­Tailed Tests About a Population Mean: σ Known s Excel Formula Worksheet A 1 Weight 2 6.04 3 5.99 4 5.92 5 6.03 6 6.01 7 5.95 8 6.09 9 6.07 10 6.07 11 5.97 12 5.96 B C Sample Size =COUNT(A2:A31) Sample Mean =AVERAGE(A2:A31) Population Std. Dev. 0.2 Hypothesized Value 6 Standard Error =C4/SQRT(C1) Test Statistic z =(C2-C5)/C7 p -Value (lower tail) =NORMSDIST(C8) p -Value (upper tail) =1-C10 p -Value (two tail) =2*(MIN(C10,C11)) Note: Rows...
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