2. Last year, the fast-food chain Chuckers in Largetown launched an advertising

Question

2. Last year, the fast-food chain Chuckers in Largetown launched an advertising cam paign featuring a popular celebrity. The CEO of Chuckers wants to determine the percent increase in sales of Chuckburgers from last year to this year. His well-paid statistician takes a simple random sample of six Chuckers restaurants in Largetown, and finds average sales of Chuckburgers per day to be as listed below. He also knows that last year's average sales of Chuckburgers per day, over all 800 Chuckers restau- rants in Largetown, was 440. Location 1 2 3 4 5 6 Last year: 586 183 231 317 311 872 This year: 692 239 235 399 381 1054 (a) Estimate the ratio of this year's Chuckburger sales in Largetown to last year's sales, and compute a bound for your error of estimation. (35 pts) (b) Is there strong evidence that the sales have increased? Explain. (10 pts)

Explanation / Answer

(a) In step 1 we will try to find the total average sales per day for last year and this year, by summing the given information across the 6 stores

i.e. Total avg sales for last year: = 2500 ; Total avg sales for this year = 3000

Now the proportion of last years total avg sales to this years total avg sales for the sample of 6 stores

= 2500/3000 = 0.8333 , this is because last years sales were less than this year

In order to compute the bound of error for this computation for the population of 800 stores, we will use test statistic

z(a/2)*(SQRT(p*(1-p)/n)) where 1-a is the confidence coefficient and SQRT denotes Square Root

So, z(a/2)*(SQRT(p*(1-p)/n)) at the 95 % confidence level = 1.96 * SQRT((0.833*(1-0.833)/6) = 0.2982

The value of z at the 95% confidence interval can be found in any standard normal distribution table = 1.96

Hence, the ratio of last year to this years sales for all the 800 stores = 0.833 (+-) 0.2982

i.e. it can be from 0.833-0.2982=0.5351 to 0.833+0.2982 = 1.131

i.e. ratio of this year sales to last years sales will be between 1/1.131 and 1/0.5351

i.e. ratio of this year sales to last years sales will be between 0.884 and 1.8681

Hence, the possibility of last years sales being greater than this year is very much there as the lower bound of the ration is lesser than 1

(b) In the second part we examine the evidence that this years sales has increased and not decreased. Here we formulate the hypothesis for testing validity of a claim

Ho : Avg sales per day this year >= 440

Ha : Avg sales per day <440

as avg sales per day over all 800 stores last year is 440 as given in the question

Since here we do not know the population standard deviation we will use the t - test statistic to validate the hypothesis

The avg sales of the sample of 6 stores this year = (692+239+235+399+381+1054)/6 = 500

and standard deviation of the same data set of 6 stores this year = 318.27

Degrees of freedom = 6-1 =5

t= (x -u)/(s/sqrt(n))

i.e. t = (500 - 440) / (318.27/SQRT(6)) = 0.461

i.e. Area in the upper tail for t = 0.461 is about 0.334 which can found by referring to a t-distribution table

Since 0.334>0.05 for the 95 % confidence interval we do not reject the null hypothesis Ho: Avg Sales this year >=440

Hence, there is strong evidence to suggest that the sales has increased from last year to this year.

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