Question

Mcdonalds waiting time from order to pick up has gotten a lot of attention in the media recently. Following is the distribution of waiting time from order to pcik up. MEDIAN=175

MEAN=180

according to qsr magazine customers are waiting longer than expected.

Group 1 studied on 50 branches. Avg waiting time = 187 seconds. standard deviation =10

Group 2 studied on 25 branches. Avg waiting time = 192 seconds. standard deviation =15

1) write the null and alternative hypothesis

2) choose one group and explain why

3) test your approach (0.07 level of significance)

4) find p value

5) show all your answers on a normal distribution graph

6) construct a 85 % confidence interval to estimate the average waiting time from order to pick up. Also, find the margin of error

Answer 1

1)

Null Hypothesis H0: Mean  waiting time from order to pick up is 180 seconds.

Alternative Hypothesis Ha: Mean waiting time from order to pick up is greater than 180 seconds.

2)

We choose Group 1 as the sample size used is large as compared with the Group 2. With large sample size, the standard error of the mean and margin of error is low.

3)

Standard error of mean = s / = 10 / = 1.4142

Test statistic, t = (Observed mean - Hypothesized Mean) / Std error

= (187 - 180) / 1.4142

= 4.95

Degree of freedom = n - 1 = 50 - 1 = 49

4)

P-value = P(t > 4.95) = 0.000005

Since p-value is less than 0.07 significance level, we reject null hypothesis H0 and conclude that there is significant evidence that mean waiting time from order to pick up is greater than 180 seconds.

5)

6)

For 85 % confidence interval , significance level = (1 - 0.85)/2 = 0.075

Critical value of t at = 0.075  and df = 49 is 1.46

Margin of error = t * Std error = 1.46 * 1.4142 = 2.064732

85 % confidence interval is,

(180 - 2.06,  180 + 2.06)

(177.94, 182.06)