In: Statistics and Probability
A) At the local pizza parlor people line up outside in their masks and proceed 1 at a time into the store to pay for and pick up the pizza. The manager who receives the orders for the pizzas wants to minimize peoples wait time. So she keeps track of the wait times for 25 people in mimutes: 40, 32, 5, 14, 7, 9, 13, 8, 9, 15, 11, 9, 15, 13, 14, 14, 7, 10, 13, 15, 18, 8, 18, 11, 7 A)Check to see if the data is normal using the normal plot.
B)If it looks normal, construct a 95% confidence bound for the mean wait time, otherwise, construct a 95% confidence bound for the median wait time.
Solution :
The given data is:
time | 40 | 32 | 5 | 14 | 7 | 9 | 13 | 8 | 9 | 15 | 11 | 9 | 15 | 13 | 14 |
14 | 7 | 10 | 13 | 15 | 18 | 8 | 18 | 11 | 7 |
1)
Plotting the histogram for the given sample data :
here, the kernel density estimate histogram plot of the given sample data does NOT appear to be normal.
Infact, the distribution is highly right-skewed.
To confirm this statistically, we can perform a statistical test,
Shapiro-test (a test of normality) using the formula :
Where,
??: Data IS Normal
??: Data IS NOT Normal
Simuliting the test on a software package, the p-value was found to be 3.64709812856745e-05 , which is less than 0.05
Hence, we reject the NULL Hypothesis at 95% level of significanceand state that the data is NOT normal.
2)
Since the data is normally NOT distributed, we create a 95 % confidence interval for the mean .
the formula for calculating the confidence interval is given by:
where,
xbar = sample mean = 13.4
t = t-cal for 95% level of significance = 1.96
s = sample standard deviation = 7.74
n = sample size = 25
so,
CI = 13.4 +- (1.96 * ( 7.74 / (25)1/2 ))
= 13.4 +- (1.96 * 1.55)
= 13.4 +- 3.03
= (16.43, 10.37)
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