Question

4. (15) The number of cups of coffee served at a local Tim Horton's during the morning rush hour between 7 and 9 AM is normally distributed. Based on sales a day is described as SLOW if less than 200 cups are served, AVERAGE 200-249 cups are sold, and BUSY if over 250 cups are sold during the day. A random sample of five days was taken and the sales figures observed. The sample mean is 208 cups and the sample standard deviation is 25.5734. a. (6) Construct a 99% confidence interval for the population average sales for this period. b. (2) Based on your confidence interval, can you conclude the type of day that the average sales would be considered as defining - SLOW, AVERAGE, or BUSY? Explain your conclusion. c. (6) Management is concerned that the average is falling below the national average of 210 cups per day. Test this hypothesis using alpha = 0.05. d. (1) What assumption must hold to allow you to construct the confidence interval and test the hypothesis in this case?

Answer 1

Solution : - Given data X = 208 S = 25.5734 n=5 (a) Now to construct a 99% confidence internal population average sales for this period Here 2=t-Confidence interval for the de t- 99 joo 2= 1-0.99 2= 0.01 t critical, Malue - ta, df = tool, ni = t...055 -1 - 10.005,4 - 4.6041

99%. I for the population average sales for this period = xtt.. 9 critical 208 + 4.60w x 25.5734 Us = 208 + ( 406041 x 11.4368) = 208 £ 52.6562 = (208-52.6562, 208+52.6562) = (155,344, 260.656) Therefre the 99% Confidence interval for the population average sales for this period is (155.344 , 260.656) (6) since the confidence Interval contains 200- 249, therefore the average Sales would be considered as "AVERAGE" (c) Now to state the null and alternative hypothesis Null bypothesis Ho: M = 210 Alternative hypothesis Hi: M <210 Here M = true average of number cups of coffee Now to find the test statistic - VS Test - statistic, t = slun 208 - 210 t= 25.5734 / 85

+ = - 11.4368 It = -0.11497 Test statistic , t= -0.1749 Now to find the degrees of freedom degrees of freedom , df = nol - df = 5-1 af = 4 Now to find the P-value P-value = 0.4348 ( By using p-value calculator for Decision a student t-Test) P- Value > 0.4348 >0.05 since the p-value is greater than significance level, so we fail to reject the null typothesis (Ho) at 0.05 level of significance. Conclusion: There is insufficient evidence to conclude that the average is falling national average of 210 cups per day. (d) This Here we assume that is normally distributed. This Construct confidence Interval number of assumption and perform Cups of coffee must hold to t-test.