Question

Let's go to the movies: A random sample of 47 Foreign Language movies made since 2000
had a mean length of 111.1 minutes, with a standard deviation of 14.6 minutes.

Part 1 of 2
Construct a 99.8% confidence interval for the true mean length of all Foreign Language movies
made since 2000. Round the answers to one decimal place.
A 99.8% confidence interval for the true mean length of all Foreign Language movies made since
2000 is __<u<___?
  

Answer 1

Answer 2

Given:

• Sample size (n): 47 movies

• Sample mean (x̄): 111.1 minutes

• Sample standard deviation (s): 14.6 minutes

• Confidence level: 99.8%

• 
  

Steps:

1. Find the critical value (z):*
   For a 99.8% confidence level, the critical value (from the z-table) is approximately 3.090. This accounts for the high confidence level, meaning we're capturing almost all of the data distribution.

2. 
   

3. Calculate the standard error (SE):
   The standard error tells us how much the sample mean might vary from the true population mean.

   $$SE=\frac{s}{\sqrt{n}}=\frac{14.6}{\sqrt{47}}\approx 2.13\text{ minutes}$$

5. Compute the margin of error (ME):
   The margin of error shows the range around the sample mean where we expect the true mean to lie.

   $$ME=z^{\ast }\times SE=3.090\times 2.13\approx 6.6\text{ minutes}$$

7. Construct the confidence interval:
   Add and subtract the margin of error from the sample mean.

   $$\text{Lower bound}=111.1-6.6=104.5\text{ minutesUpper bound}=111.1+6.6=117.7\text{ minutes}$$

 Answer is :

A 99.8% confidence interval for the true mean length of all Foreign Language movies made since 2000 is 104.5 < μ < 117.7 minutes