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Question 9 of 20 1.0/ 1.0 Points Click to see additional instructions A restaurant chain 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 10 of 20 1.0/ 1.0 Points A high school offers math placement exams for incoming freshmen to place students into the appropriate math class during their freshman year. Three different middle schools were sampled and the following pass/fail results were found. Run a test for independence at the 0.10 level of significance. School A School B School C Pass 40 33 50 Fail 59 45 67 After running an independence test, can it be concluded that pass/fail rates are dependent on school?
B. Yes, it can be concluded that pass/fail rates are dependent on school because the p-value = 0.9373. C. Yes, it can be concluded that pass/fail rates are dependent on school because the p-value = 0.0627. D. No, it cannot be concluded that pass/fail rates are dependent on school because the p- value = 0.9373. Answer Key: D 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. We need to calculate the Expected Counts. Then sum up the rows and column. You need to find the probability of the row and then multiple it by the column total. School A School B School C Sum Pass 40 33 50 123 Fail 59 45 67 171 Sum 99 78 117 294 School A School B School C Pass =99*(123/294)=78*(123/294) =117*(123/294) Fail =99*(171/294)=78*(171/294) =117*(171/294) 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) = 0.9373 0.9373 > .10, Do Not Reject Ho. No, it cannot be concluded that pass/fail rates are dependent on school.
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