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MC Qu. 98 Here is an Excel ANOVA table... Here is an Excel ANOVA table that summarizes the results of an experiment to assess the effects of ambient noise level and plant location on worker productivity. The test used a = .05. Source of Variation 55 df MS F P-value F crit Plant location 3. 0075 1. 0025 2.561 0. 1199 3. 862 W W Noise level 8. 4075 2. 8025 7.160 0. 0093 3.863 Error 3.5225 0. 3914 Total 14.9375 s the effect of noise level significant at a = .05? Multiple Choice O Yes O No O We need more information to say.MVC A proofreader ads Tor... A proofreader checked 160 ads for grammatical errors. The sample frequency distribution is shown below. Number of Errors 1 2 3 Observed Frequency 10 65 71 14 A goodness-of-fit test to determine whether this sample came from a uniform distribution would result in a chi-square test statistic of approximately Multiple Choice 55 O 79 O 85 O 161MC Qu. 51 A proofreader checked 160 ads for... A proofreader checked 160 ads for grammatical errors. The distribution obtained is shown below. Number of Errors 1 2 3 Observed Frequency 10 65 71 14 At a = .01, what decision would we reach in a goodness-of-fit test to see whether this sample came from a uniform distribution? Multiple Choice O Reject the null and conclude the distribution is not uniform. O Conclude that there is insufficient evidence to reject the null. O No conclusion can be made due to small expected frequencies. O No conclusion can be made due to inadequate sample size.Section Exercise 15-35 Oxnard Kortholt, Lid., employs 50 workers. Research question: At a = .05, do Oxnard employees differ significantly from the national percent distribution? Oxnard Employees Health Care Visits National Percentage Frequency No visits 15.0 3 1-3 visits 42.4 21 4-9 visits 22.7 18 10 or more visits 19.9 8 Total 100.0 50 Expicture Click here for the Excel Data File (a) The hypothesis for the given issue is He: Oxnard employees do not differ significantly from the national percent distribution. O Yes O No (b) Calculate the chi-square test statistic, degrees of freedom and the p-value. (Round your test statistic value to 3 decimal places and the p-value to 4 decimal places.) Test statistic d.f. p-value (c) We fail to reject the null hypothesis. O Yes O NoI... Kortholts that fail to meet certain precise specifications must be reworked on the next day until they are within the desired specifications. A sample of one day's output of kortholts from the Melodic Kortholt Company showed the frequencies shown below. Plant A Plant B Row Total Specification Met 185 85 270 Specification Not Met 15 15 30 Col Total 206 100 300 Find the chi-square test statistic for a hypothesis of independence. Multiple Choice O 7.22 O 4.17 O 5.13 O 6.08MC Qu. 59 Kortholts that fail to meet certain... Kortholts that fail to meet certain precise specifications must be reworked on the next day until they are within the desired specifications. A sample of one day's output of kortholts from the Melodic Kortholt Company showed the frequencies shown below. Plant A Plant B Row Total Specification Met 185 85 270 Specification Not Met 15 15 30 Col Total 200 100 300 Find the p-value for the chi-square test statistic for a hypothesis of independence. Multiple Choice O Less than .01 O Between .01 and .025 O Between .025 and .05 O Greater than .05u want to sell your house, and you decide to obtain an appraisal on it. Looking at past data, you discover that actual prices obtained for houses and the appraisals given for them prior to their sale were as shown low. Actual Selling Price Appraisal Up to $500,000 $500, 001 or more Row Total Up to $500, 000 21 9 30 $500, 001 or more 14 20 Col Tot 35 15 50 ased on these data we can say that Multiple Choice O no conclusion is possible without knowing a. O appraisal and actual price are not independent at a = .05. O appraisal and actual price are independent at any a. O the degrees of freedom are insufficient for a decision