2. The owner of a chain of fast-food restaurants is fanatically intent on standa

Question

2. The owner of a chain of fast-food restaurants is fanatically intent on standardization. Every week she selects an aspect of the operation, and examines data pertaining to it. This week she has decided to concentrate on the size of the serving of mashed potatoes in the Bang-O-Mash special Here are some data Quebec Ontario 90 55.5 49.2 Sample size Mean size of serving (grams) Standard deviation 10.9 (grams) 150 6.0 a) Test whether the mean serving in Quebec is significantly greater than 50 g. Use a 10% significance level b) The mean serving in Quebec is rumoured to be larger than the mean serving in Ontario. Test whether the mean serving is significantly larger. Use a 5% level of significance. c) Bang-O-Mash has not been successful in all outlets. A sample of 200 outlets in Ontario included 30 in which Bang-O-Mash was considered successful. In Quebec, a sample of 150 outlets included 27 in which Bang-O- Mash was considered successful. Test at the 1% level whether the proportions of outlets where Bang-O-Mash is considered successful are significantly different in Quebec and Ontario. What is the p-value for this test?

Explanation / Answer

(a) Here

sample mean x = 55.5 gm

Sample standard deviation s = 10.9

sample size n = 90

standard error of sample mean se0 = s/sqrt(n) = 10.9/sqrt(90) = 1.149

Test statistic

t = (55.5 - 50)/1.149 = 4.787

Here dF = 90 - 1 = 89 ; alpha = 0.10

tcritical = 1.2911 [one tailed test]

here t > tcritical so we can say that mean serving in Quebec is significantly greater 50 g.

(b) Pooled standard deviation

sp = sqrt [{(n1-1)s12+ (n2 -1)s22}/(n1 + n2-2)] = sqrt [(89 * 10.92 + 149 * 62)/(89 + 149)] = sqrt(66.97) = 8.183

standard error of difference of sample mean se0 = sp * sqrt (1/n1 + 1/n2) = 8.183 * sqrt (1/90 + 1/150) = 1.0911

Test statistic

t = (x1 -x2)/se0 = (55.5 - 49.2)/ 1.0911 = 5.774

Here dF = 90 + 150 -2 = 238

tcritical = t238,0.05 = 1.6513

here t > tcritical so we can say that mean serving in Quebec is significantly greater than mean serving in ontario.

(c) Here,

pontario = p1 = 30/200 = 0.15

pQuebec = p2 = 27/150 = 0.18

Pooled estimate p = (30 + 27)/(200 + 150) = 0.163

standard error of pooled estimate sep = sqrt [p* (1-p) * (1/n1 + 1/n2)] = sqrt [0.163 * 0.837 * (1/150 + 1/200)] = 0.0399

Here

Test statistic

Z = (pQuebec - pontario)/se0 = (0.18 - 0.15)/ 0.04 = 0.75

Here the test is two tailed.

so p - value = 2 * Pr(Z > 0.75) = 2 * 0.2267 = 0.452

so here as p value is greater than significance level 0.01 so we can say that proportions of outlets where Bang- O - mash is considered successful are not significantly different in Quebec and Ontario.

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