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Question 11 of 20 1.0/ 1.0 Points Click to see additional instructions A manufacturing company knows that their machines produce parts that are defective on occasion. They have 4 machines producing parts, and want to test if defective parts are dependent on the machine that produced it. They take a random sample of 321 parts and find the following results. Test at the 0.05 level of significance. Machine 1 Machine 2 Machine 3 Machine 4
Defective 10 15 16 9 Non-Defective 72 75 66 58 Enter the test statistic - round to 4 decimal places. 1.9943 Answer Key: 1.9943 Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. Machine 1 Machine 2 Machine 3 Machine 4 Sum Defective 10 15 16 9 50 Non-Defective 72 75 66 58 271 Sum 82 90 82 67 321 Machine 1 Machine 2 Machine 3 Machine 4 Defective =82*(50/321) =90*(50/321) =82*(50/321) =67*(50/321) Non-Defective =82*(271/321) =90*(271/321) =82*(271/321) =67*(271/321) Now that we calculated the Expected Counts we need to find the Test Statistic. Test Stat = You will need to all 8 Count values but I am only showing you 3 because there isn't room to write out the entire equation. Question 12 of 20 1.0/ 1.0 Points Click to see additional instructions A restaurant that has 3 locations in Portland is trying to determine which of their 3 locations they should keep open on New Year’s Eve. They survey a random sample of customers at each location and ask each whether or not they plan on going out to eat on New Year’s Eve. The results are below. Run a test for independence to decide if the proportion of customers that will go out to eat on New Year’s Eve is dependent on location. Use α=0.05. Enter the P -Value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. NW Location NE Location SE Location Will Go Out 66 40 45 Won’t Go Out 20 25 20
Feedback: We are running a Chi-Square Test for Independence. Copy and Paste the table into Excel. You are given the Observed Counts in the table. Next you need to sum the rows and columns. Once you have those you need to calculate the Expected Counts. You need to find the probability of the row and then multiple it by the column total. NW Location NE Location SE Location Sum Will Go Out 66 40 45 151 Won’t Go Out 20 25 20 65
Sum 86 65 65 216 NW Location NE Location SE Location Will Go Out =86*(151/216) =65*(151/216) =65*(151/216) Won’t Go Out =86*(65/216) =65*(65/216) =65*(65/216) Now that we calculated the Expected Count we can use Excel to find the p-value. Use =CHISQ.TEST(highlight actual counts, highlight expected counts) = 0.1294 Question 13 of 20 1.0/ 1.0 Points Click to see additional instructions A university changed to a new learning management system during the past school year. The school wants to find out how it’s working for the different departments – the results in preference found from a survey are below. Run a test for independence at α=0.05.
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