A restaurant chain that has 3 locations in Portland is trying to determine

# A restaurant chain that has 3 locations in portland

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Question 15 of 201.0/ 1.0 PointsA 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 customersat 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=0.05Can it be concluded that the choice to go out on New Year's Eve is dependent on restaurant location? NW LocationNE LocationSE LocationWill Go Out664045Won’t Go Out202520 . A. No, it cannot be concluded that the choice to go out on New Year's Eve is dependent on restaurant location because the p-value = 0.8706
B. Yes, it can be concluded that the choice to go out on New Year's Eve is dependent on restaurant location because the p-value = 0.8706C. Yes, it can be concluded that the choice to go out on New Year's Eve is dependent on restaurant location because the p-value = 0.1294.D. No, it cannot be concluded that the choice to go out on New Year's Eve is dependent on restaurant location because the p-value = 0.1294.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. 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) = 0.12940.1294 > 0.05, Do Not Reject Ho. No, it cannot be concluded that the choice to go out on New Year's Eve is dependent on restaurant location Question 16 of 201.0/ 1.0 PointsClick to see additional instructionsAn electronics store has 4 branches in a large city. They are curious if sales in any particular department are different depending on location. They take a random sample of 4 purchases throughout the 4 branches – the results are recorded below. Run an independence test for the data below at the 0.05 level of significance.AppliancesTVComputersCell PhonesBranch 156286324Branch 244225527Branch 353174933Branch 451316629Enter the P-Value - round to 4 decimal places. Make sure you put a 0 in front of the decimal. 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. Appliances TV Computers Cell Phones Branch 1 56 28 63 24 Branch 2 44 22 55 27 Branch 3 53 17 49 33 Branch 4 51 31 66 29 Sum 204 98 233 113 Appliances TV