Question

A movie theater manager wants to determine whether popcorn sales have increased since the theater switched from using "butter-flavored topping" to real butter. Historically the average popcorn revenue per weekend day was approximately $3,500. After the theater started using real butter, the manager randomly sampled 12 weekend days and calculated the sample’s summary statistics. The average revenue per weekend day in the sample was approximately $4,200 with a standard deviation of $140. Select the function that would correctly calculate the 90% range of likely sample means.A. 3,500±CONFIDENCE.T(0.10,140,12)B. 4,200±CONFIDENCE.T(0.10,140,12)C. 3,500±CONFIDENCE.NORM(0.10,140,12)D. 4,200±CONFIDENCE.NORM(0.10,140,12)

Answer 1

Answer:

B. 4,200±CONFIDENCE.T(0.10,140,12)

Step-by-step explanation:

We are in posession's of the sample standard deviation, so the t-distribution is used.

The confidence interval is a function of the sample mean and the margin of error.

That is:

$C.I = S_{M} \pm M_{E}$

In which $S_{M}$ is the sample mean, and the $M_{E}$ is the margin of error, related to the confidence level, the sample's standard deviation and the sample size.

So the correct answer is:

B. 4,200±CONFIDENCE.T(0.10,140,12)

Answer 2

Answer:

B. 4,200±CONFIDENCE.T(0.10,140,12)

Step-by-step explanation:

The sample mean revenue per weekend day in the sample = x = $4200

The standard deviation of the revenue= δ=$140

The sample size , n= 12 weekend days

z*=will represent the appropriate z* value according to 90% confidence level =1.645

The formula to apply is;

x ± z* δ/√n

z*δ= 1.645×140 =230.3

√n = √12 =3.464

230.3/3.464 =66.48

C.I= $4200 ± 66.48

Answer

B. 4,200±CONFIDENCE.T(0.10,140,12).