1. suppose we wish to predict average customer spending. we take a small sample of size 50 and find the sample mean to be $200/mo. with a sample standard deviation of $10/mo. place a 99% confidence interval around the population average spending.

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Yogeshkumarigo Yogeshkumarigo · Answer 1

The average customers are spending in the range of ( 196.36 , 203.64) for the given mean and standard deviation.

As given in the question,

Sample size 'n' = 50

Sample mean 'μ ' is equal to 200

Standard deviation 'σ' = 10

z-score for 99% Confidence interval = 2.576

Confidence interval = mean ± z ( σ / √n )

Substitute the values we get,

Confidence Interval = 200 ± 2.576( 10 / √50)

= 200± 2.576 ( 1.414)

= 200 ± 3.642464

Lower bound of the required confidence interval = 200 - 3.642464

= 196.36

Upper bound of the required confidence interval = 200 + 3.642464

= 203.64

Therefore, the average customer is spending between ( 196.36 , 203.64) range from the given mean and standard deviation.

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