A hair salon in Cambridge, Massachusetts, reports that on seven randomly selecte

Question

A hair salon in Cambridge, Massachusetts, reports that on seven randomly selected weekdays, the number of customers who visited the salon were 58, 42, 43, 38, 55, 28, and 39. It can be assumed that weekday customer visits follow a normal distribution. Use Table 2.

Construct a 90% confidence interval for the average number of customers who visit the salon on weekdays. (Round intermediate calculations to 4 decimal places, "sample mean" and "sample standard deviation" to 2 decimal places and "t" value to 3 decimal places, and final answers to 2 decimal places.)

Construct a 99% confidence interval for the average number of customers who visit the salon on weekdays. (Round intermediate calculations to 4 decimal places, "sample mean" and "sample standard deviation" to 2 decimal places and "t" value to 3 decimal places, and final answers to 2 decimal places.)

What happens to the width of the interval as the confidence level increases?

Construct a 90% confidence interval for the average number of customers who visit the salon on weekdays. (Round intermediate calculations to 4 decimal places, "sample mean" and "sample standard deviation" to 2 decimal places and "t" value to 3 decimal places, and final answers to 2 decimal places.)

Explanation / Answer

a)
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=43.28
Standard deviation( sd )=10.291
Sample Size(n)=7
Confidence Interval = [ 43.28 ± t a/2 ( 10.291/ Sqrt ( 7) ) ]
= [ 43.28 - 1.9432 * (3.89) , 43.28 + 1.9432 * (3.89) ]
= [ 35.72,50.84 ]
b)

Confidence Interval = [ 43.28 ± t a/2 ( 10.291/ Sqrt ( 7) ) ]
= [ 43.28 - 3.7074 * (3.89) , 43.28 + 3.7074 * (3.89) ]
= [ 28.86,57.7 ]

c)
As the confidence level increases, the interval becomes wider and less precise.

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