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- The manager of a local record store is interested in the effectiveness of an advertisingcampaign on sales. They randomly select 10 popular CDs (all in the Billboard top 100),and place an ad in the local school newspaper saying that the store has these (and more)in stock. They measure the sales of these 10 CDs for 1 week prior to the ad, and then for1 week after the ad. The number of CDs sold each week is presented below. Was there anincrease in the number of CDs sold after the ad? Conduct a 6-step hypothesis test.CD Numbersold beforeadNumbersold after adA 25 32B 18 24C 3 7D 42 40E 16 19F 20 25G 23 23H 32 35I 60 65J 40 43 Please solve using 6 steps hypothesis and solve in excelfind each value requested for the set scores listed below a. n b. Ex c. Ex^2 d. 10+E(x-3)^2 5,4,6,1,4,5,2,6the comic book store owner is curious in how well radio and newspaper advertisements work. A sample of 22 counties with almost equal populations was gathered during the month. For each of the counties, the owner has set aside a particular sum for radio and newspaper advertising. Additionally, the owner was able to access comic book sales information for the stores he controlled in each county and classified it as Low or High. COUNTY (X1) Comic book Sales (X2) (0 = Low, 1 = High) Radio Ads (X3) (in thousands) Newspaper Ads (X4) (in thousands) 1 1 0 40 2 0 0 40 3 0 25 25 4 0 25 25 5 0 30 30 6 0 30 30 7 0 35 35 8 0 35 35 9 0 40 25 10 0 40 25 11 1 45 45 12 1 45 45 13 1 50 0 14 1 50 0 15 1 55 25 16 1 55 25 17 1 60 30 18 1 60 30 19 1 65 35 20 1 65 35 21 1 70 40 22 1…
- A study was undertaken to determine whether the consumers have any preference among four brands of cars. A sample of 200 people was used to collect data. I. II. Brand Number of consumers A 49- B B 52- α с 61+ B D 38+a Chi-square goodness of fit test can be applied to the above data set. Write 2 reasons. Using chi squared goodness of fit test, determine whether the consumers have any preference among four brands of cars. State your hypotheses and all workings clearly. (Critical value for degrees of freedom 3 and significance level 0.05 is 7.815.)The scores of a test for an engineering class of 30 students are shown here. (Due to the nature of this problem, do not use rounded intermediate values in your calculations-including answers submitted in WebAssign.) Scores: 81, 54, 55, 56, 96, 86, 98, 61, 90, 52, 92, 76, 66, 84, 79, 70, 71, 88, 62, 74, 53, 65, 55, 78, 79, 58, 60, 72, 51,80 (a) Determine the frequency (f), midpoint (x), and xf for each range and fill in the table below. (First, organize the scores into the following ranges: 50-59, 60-69, 70-79, 80-89, and 90-99.) (b) Using the following equation, determine the mean value (x) and fill in the table below. Σ(XA) n X = (c) Calculate (x − x) for each range and fill in the table below. (d) Calculate (x - x)²f for each range and fill in the table below. Range Frequency f Midpoint x 50-59 60-69 70-79 80-89 90-99 S = Enter a number. S = (e) Using the following equation, compute the standard deviation of the class scores. Σ(x-x) ²f n - 1 Σ(x-x) ²4 n - 1 = xf 29 X X - X (x - x)²fThe following table was generated from the sample data of 1010 college students regarding the number of parking tickets the student receives in a semester, the student's age, and the student's GPA. The dependent variable is the student's GPA, the first independent variable (x1�1) is the number of parking tickets, and the second independent variable (x2�2) is the student's age. Coefficients Standard Error t-Stat p-value Intercept −18.593656−18.593656 3.6727623.672762 −5.062581−5.062581 0.0038910.003891 Number of Parking Tickets 0.3232630.323263 0.0657790.065779 4.9144144.914414 0.0044190.004419 Student's Age 1.0181271.018127 0.1791910.179191 5.6817895.681789 0.0023530.002353 Copy Data Step 1 of 2 : Write the multiple regression equation for the computer output given. Round your answers to three decimal places
- Comment on the P-value for testing the claim p1- p2=0The pass result of fifty students who took up a class test is given below: Marks Number of Students 4 8 5 10 9. 7 4 9. 3 If the average marks for all the fifty students were 5.16, find out the average marks of ti students who failed.We are interested in estimating the proportion of people that can identify Nebraska but not Vermont. In a class of 55 students, each student was requested to select a sample of 6 people, show each person a map of the US and ask the person to identify the states of Vermont and Nebraska. The following numbers are the counts that each student got, that is, each number is the count of people in the student sample of 6, that could identify Nebraska but not Vermont.3344010110020030002020001100002010000001000200000001010Note: You can cut and past these numbers into an excel spreadsheet to help you calculate the mean and the standard deviation and compute the confidence interval. If you have trouble pasting them, paste them to a text file first and then copy from there.When we collect all the students' data, we have a sample size of 6x55=330.Of the sample of 330 people, what proportion were able to locate Nebraska but not Vermont? (round answer to 3 decimals) Find a 95% confidence interval for…
- hand written plzA student takes a true-false test that has 13 questions and guesses randomly at each answer. Let X be the number of questions answered correctly. Find P(11 or more)A magazine collected the ratings for food, decor, service and the cost per person for a sample of 96 restaurants. They combined the ratings to create a summated rating and used that to predict the cost of a restaurant meal. The data are modeled by Y; = - 30.871 + 0.9349X;, where X; is the summated ratings and Y; is meal cost. For these data, Syx = 8.575, X= 61.02, and h; = 0.027122 when X = 52. Complete parts (a) through (c). %3D %3D a. Construct a 90% confidence interval estimate of the mean cost of a meal for restaurants that have a summated rating of 52. |SHYIX =x, 5 (Type integers or decimals. Round to three decimal places as needed. Use ascending order.)