Customers arrive at a movie theater at the advertised movie time only to find that they have to sit through several previews and preview ads before the movie starts. Many complain that the time devoted to previews is too long (The Wall Street Journal, October 12, 2012). A preliminary sample conducted by The Wall Street Journal showed that the standard deviation of the amount of time devoted to previews was four minutes. Use that as a planning value for the standard deviation in answering the following questions. If we want to estimate the population mean time for previews at movie theaters with a margin of error of 75 seconds, what sample size should be used

the sample size should be 39

Step-by-step explanation:

The sample mean is the point estimate for the population mean. Confidence interval is written as

Sample mean ± margin of error

Margin of error = z × σ/√n

Where

σ = population standard Deviation

n = number of samples

z represents the z score corresponding to the confidence level

From the information given,

σ = 4 minutes

Margin of error = 75 seconds. Converting to minutes, it becomes 75/60 = 1.25 minutes

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, confidence level of 95% is 1.96

Therefore,

1.25 = 1.96 × 4/√n

1.25/1.96 = 4/√n

0.6378 = 4/√n

√n = 4/0.6378 = 6.27

n = 6.27² = 39

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.

answer: the answer is 110