"Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 73 manufacturing companies located in the Southwest. The mean expense is $50.27 million and the standard deviation is $10.93 million. Is it reasonable to conclude the sample data are from a population that follows a normal probability distribution? Advertising Expense ($ Million) Number of Companies 25 up to 35 35 up to 45 45 up to 55 55 up to 65 65 up to 75 18 29 12 Total 73 Click here for the Excel Data File State the decision rule. Use the 0.10 significance level. (Round your answer to 2 decimal places.) Ho: The population of advertising expenses follows a normal distribution. H: The population of advertising expenses does not follow a normal distribution. Reject H0 if chi-square > Compute the value of chi-square. (Round your answer to 2 decimal places.) Chi-square value What is your decision regarding Hg? H0. This data from a normal distribution. could be The Damon family owns a large grape vineyard in western New York along Lake Erie. The grapevines must be sprayed at the beginning of the growing season to protect against various insects and diseases. Two new insecticides have just been marketed: Pernod 5 and Action. To test their effectiveness, three long rows were selected and sprayed with Pernod 5, and three others were sprayed with Action. When the grapes ripened, 440 of the vines treated with Pernod 5 were checked for infestation. Likewise, a sample of 360 vines sprayed with Action were checked. The results are: Number of Vines Checked (sample size) Insecticide Number of Infested Vines Pernod 5 440 30 Action 360 36 At the 0.02 significance level, can we conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action? Hint. For the calculations, assume the Pernod 5 as the first sample. 1. State the decision rule. (Negative amounts should be indicated by a minus sign. Do not round the intermediate values. Round your answers to 2 decimal places.) H0 is rejected if z < or z> b. Compute the pooled proportion. (Do not round the intermediate values. Round your answer to 2 decimal places.) Pooled proportion c. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Do not round the intermediate values. Round your answer to 2 decimal places.) Value of the test statistic

"Advertising expenses are a significant component of the cost of goods sold. Listed below is a frequency distribution showing the advertising expenditures for 73 manufacturing companies located in the Southwest. The mean expense is $50.27 million and the standard deviation is $10.93 million. Is it reasonable to conclude the sample data are from a population that follows a normal probability distribution? Advertising Expense ($ Million) Number of Companies 25 up to 35 35 up to 45 45 up to 55 55 up to 65 65 up to 75 18 29 12 Total 73 Click here for the Excel Data File State the decision rule. Use the 0.10 significance level. (Round your answer to 2 decimal places.) Ho: The population of advertising expenses follows a normal distribution. H: The population of advertising expenses does not follow a normal distribution. Reject H0 if chi-square > Compute the value of chi-square. (Round your answer to 2 decimal places.) Chi-square value What is your decision regarding Hg? H0. This data from a normal distribution. could be The Damon family owns a large grape vineyard in western New York along Lake Erie. The grapevines must be sprayed at the beginning of the growing season to protect against various insects and diseases. Two new insecticides have just been marketed: Pernod 5 and Action. To test their effectiveness, three long rows were selected and sprayed with Pernod 5, and three others were sprayed with Action. When the grapes ripened, 440 of the vines treated with Pernod 5 were checked for infestation. Likewise, a sample of 360 vines sprayed with Action were checked. The results are: Number of Vines Checked (sample size) Insecticide Number of Infested Vines Pernod 5 440 30 Action 360 36 At the 0.02 significance level, can we conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action? Hint. For the calculations, assume the Pernod 5 as the first sample. 1. State the decision rule. (Negative amounts should be indicated by a minus sign. Do not round the intermediate values. Round your answers to 2 decimal places.) H0 is rejected if z < or z> b. Compute the pooled proportion. (Do not round the intermediate values. Round your answer to 2 decimal places.) Pooled proportion c. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Do not round the intermediate values. Round your answer to 2 decimal places.) Value of the test statistic

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.4: Distributions Of Data

Problem 19PFA

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