Question

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%

Answer 1

Answer: 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%

Answer 2

Answer:

The claim can be made with 86.1% level of confidence.

Step-by-step explanation:

We are given that 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.

Let $\bar X$ = mean amount spent on Halloween costumes for all U.S. adults aged 18–24

Assume data follows normal distribution.

So, the z score probability distribution is given by;

         Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean amount spent = $37.51

            $\sigma$ = standard deviation = $16.44

             n = sample of U.S. adults = 100

So, probability that the the mean amount spent on Halloween costumes for all U.S. adults aged 18–24 is between $35.07 and $39.95 is given by = P($35.07 < $\bar X$ < $39.95) = P($\bar X$ < $39.95) - P($\bar X$ $\leq$ $35.07)

   P($\bar X$ < 39.95) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{39.95-37.51}{\frac{16.44}{\sqrt{100} } }$ ) = P(Z < 1.48) = 0.93056                                P($\bar X$ $\leq$ 35.07) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{35.07-37.51}{\frac{16.44}{\sqrt{100} } }$ ) = P(Z $\leq$ -1.48) = 1 - P(Z < 1.48)

                                                              = 1 - 0.93056 = 0.06944

Therefore, P($35.07 < $\bar X$ < $39.95) = 0.93056 - 0.06944 = 0.861

Hence, with 86.1% level of confidence this claim can be made.