Question

1. 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.

1. (6) Construct a 99% confidence interval for the population average sales for this period.
2. (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.
3. (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.
4. (1) What assumption must hold to allow you to construct the confidence interval and test the hypothesis in this case?

Answer 1

Construct a 99% confidence interval for the population average sales for this period.

We are given sample mean = ji= 208 sample standard deviation = 0 = 25.5734 n = 5 a/2 = (1-0.99)/2= 0.005 df = n - 1 = 4 Using the Student t-table, we find the tstat for a/2 = 0.005 and df = 4 tstat = 4.6040949 Margin of error = ME = tstat x = 4.6040949 x 25.5734 = 52.6559841 Confidence interval = u Etstat x = 208+ 4.6040949 x 25.5789 = 208 + 52.6559841 Therefore the 99% confidence interval for the mean is (155.344.260.656)

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.

The confidence interval extends from 155 cups to 260 cups. Hence if would fall in a slow to an average type of day

.Management is concerned that the average is falling below the national average of 210 cups per day. Test this hypothesis using alpha = 0.05.

The provided sample mean is X = 208 and the sample standard deviation is s = 25.5734, and the sample size is n = 5. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ha: u<210 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a left- tailed test is te = -2.132. The rejection region for this left-tailed test is R=t:t < -2.132 (3) Test Statistics The t-statistic is computed as follows: X - MO s/n 208-210 25.5734/ 75 (4) Decision about the null hypothesis = -2.132, it is then concluded that the null hypothesis is Since it is observed that t = -0.175 > t not rejected. Using the P-value approach: The p-value is p = 0.4348, and since p=0.4348 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean is less than 210, at the 0.05 significance level.

Hence we can conclude that there is not enough evidence to conclude that the average is less than the national average of 210

What assumption must hold to allow you to construct the confidence interval and test the hypothesis in this case?

THe sample must be taken randomly.
Each sample must be independent