A hair salon in Cambridge Massachusetts reports that on seve

A hair salon in Cambridge, Massachusetts, reports that on seven randomly selected weekdays, the number of customers who visited the salon were 89, 47, 48, 22, 40, 38, and 90. 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?

Solution

Confidence Interval For t- Single Mean

CI = x ± t_a/2 * (sd/ Sqrt(n))

Where, x = Mean

sd = Standard Deviation

a = 1 - (Confidence Level/100)

t_a/2 = t-table value

CI = Confidence Interval

Mean(x)=53.42

Standard deviation( sd )=24.14

Sample Size(n)=7

(a)

For 90% a=0.1

t_0.05=2.015

Confidence Interval = [ 53.42 ± 2.015 ( 24.14/ Sqrt ( 7) ) ] =

[ 53.42 - 2.015*9.124 , 53.42 + 2.015*9.124 ] = [ 35.04 , 71.80 ]

(b)

For 99% a=0.01

t_0.005=4.032

Confidence Interval = [ 53.42 ± 2.015 ( 24.14/ Sqrt ( 7) ) ] =

[ 53.42 - 4.032*9.124 , 53.42 + 4.032*9.124  ] = [ 16.63 , 90.20 ]

(c)

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