Task 1. Sample size = 100. Average spending = $5.46. St

Task 1
Sample size = 100
Average spending = $5.46
Standard deviation of population = $2.47
a) 99% confidence interval
The z value for 99% CI = 2.58
b) Sample size =?
Margin of error = $.50
Confidence interval = 95%
The z value = 1.96
Sample size would be 94 customers.
c) Hypothesis testing
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It is a right tailed hypothesis testing and thus, the p value for this is 0.014408.
Significance level = 0.01
Clearly, the p value is exceeding than significance level and thus, we fail to reject the null
hypothesis. Therefore, it can be said that average spending was $6.00 per visit.
Task 2
Sample size = 20
Average age = 45.23 years
Standard deviation of sample = 20.67
a) 90% confidence interval
Degree of freedom = 20-1 = 19
The t value for 90% CI = 1.73
It can say with 90% confidence that the average age would fall between 37.2340 years and
53.2260 years.
b) Margin of error
Margin of error is t value multiply by standard error.
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