Ahair salon in cambridge, massachusetts, reports that on seven randomly selected weekdays, the number of customers who visited the salon were 40, 30, 28, 22, 36, 16, and 50. it can be assumed that weekday customer visits follow a normal distribution. [you may find it useful to reference the t table.] a. construct the 90% confidence interval for the average number of customers who visit the salon on weekdays. (round intermediate calculations to at least 4 decimal places, "sample mean" and "sample standard deviation" to 2 decimal places. round "t" value to 3 decimal places and final answers to 2 decimal places.)

a: 28 < µ < 34

Step-by-step explanation:

We need the mean, var, and standard deviation for the data set.  See first attached photo for calculations for these...

We get a mean of 222/7 = 31.7143

and a sample standard deviation of: 4.3079

We can now construct our confidence interval.  See the second attached photo for the construction steps.

They want a 90% confidence interval.  Our sample size is 7, so since n < 30, we will use a t-score.  Look up the value under the 10% area in 2 tails column, and degree of freedom is 6 (degree of freedom is always 1 less than sample size for confidence intervals when n < 30)

The t-value is: 1.943

We rounded down to the nearest person in the interval because we don't want to over estimate.  It said 28.55, so more than 28 but not quite 29, so if we use 29 as the lower limit, we could over estimate.  It's better to use 28 and underestimate a little when considering customer flow.

answer: -0.5 would be the most likley outcome

step-by-step explanation:

for the free

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