Question

Include Matlab question，other write by hand，thanks

A data engineer working for The Gabba Stadium has recorded the number of people attending AFL events in 2018. This data has been provided on Blackboard in a.csv file (GabbaAttendance.csv). The data engineer claims that the average attendance at an event is 15,000 people and your task is to determine whether this is valid or not, by following the steps below 1. Using MATLAB, import the data using csvread and determine the number of samples, sample mean and standard deviation (using in-built MATLAB functions) 2, Determine the 95% confidence interval for the number of people attending an AFL event at The Gabba Stadium. Interpret this interval. 3. Conduct a hypothesis test to see if there is sufficient evidence to suggest that the data engineers claim is incorrect (use α = 0.05). Use the comparison between Twa and 1x-1,1-e2 for your hypothesis test. . Within the dataset there is one event which only recorded an attendance of 6060. Exclude this observation and re-calculate the confidence interval and hypothesis test to determine if you get a different result to your answer in 3

Answer 1

1)

From the given data, we calculate:

n = 10

Mean (xbar) = 17529.4

Std Dev (s) = 5,081.91

Std Error (SE) = s/n^1/2 = 1,607.04

2)

At alpha = 0.05,

Z_Critical = 1.96

Hence,

95% CI = Mean +/- Z_Critical *SE = 17529.4 +/- 1.96*1,607.04 = {14379.6,20679.2}

Interpretation

There is 95% probability that true mean lies in the interval {14379.6,20679.2}

3)

Alpha = 0.05

Null and Alternate Hypothesis

H₀: µ = 15000

H_a: µ <> 15000

Test Statistic

t = (xbar - µ₀)/ SE = (17529.4-15000)/ 1,607.04 = 1.57

p-value = TDIST(1.57,9,2) = 0.1499

Since the p-value is greater than 0.05, hence we fail to reject the null hypothesis

4)

When we exclude the outlier,

n = 9

Mean (xbar) = 18803.78

Std Dev (s) = 3283.84

Std Error (SE) = s/n^1/2 = 1094.613

95% CI = Mean +/- Z_Critical *SE = 18803.78 +/- 1.96*1094.613 = {16,658.34, 20,949.22}

Alpha = 0.05

Null and Alternate Hypothesis

H₀: µ = 15000

H_a: µ <> 15000

Test Statistic

t = (xbar - µ₀)/ SE = (18803.78-15000)/ 1094.613 = 3.47

p-value = TDIST(3.47,9,2) = 0.006993

Since the p-value is less than 0.05, hence we reject the null hypothesis.