In a sample of 100 U. S. adults aged 18–24 who celebrate Halloween, the mean amount spent on a costume was $37.51 with a standard deviation of $16.44. A retail specialist claims that the mean amount spent on Halloween costumes for all U. S. adults aged 18–24 is between $35.07 and $39.95. With what level of confidence can this claim be made? Answer is 85.9%

The level of confidence is 86%.

Step-by-step explanation:

From the information given

Number of sample, n = 100

Mean, u = $37.51

Standard deviation, s = $16.44

Firstly, we would determine the upper and lower limits of the mean. Therefore,

For upper limit, it is

39.95 - 37.51 = 2.44

For lower limit, it is

37.51 - 35.07 = 2.44

Limit = ± 2.44

We will apply the formula

Confidence interval which is expressed as

= mean ± z × standard deviation/√n

It becomes

37.51 ± z × 16.44/√100

= 37.51 ± z × 1.644

= 37.51 ± 1.644z

Therefore,

1.644z = 2.44

z = 2.44/1.644 = 1.48

The level of confidence corresponding to the z value is 86%

i believe the correct answer is b.