1.The average number of acts of physical violence in a random sample of 150 TV movies was 4.5 acts with a standard deviation of 0.75. At the 95% confidence interval level, what is your estimation of the mean range of all movies shown in the television? Explain the result in one sentence.

Question

1.The average number of acts of physical violence in a random sample of 150 TV movies was 4.5 acts with a standard deviation of 0.75. At the 95% confidence interval level, what is your estimation of the mean range of all movies shown in the television? Explain the result in one sentence.

Answer 1

95% = mean ± 1.96 SD

Insert values and calculate.

Answer 2

To estimate the mean range of all movies shown on television at a 95% confidence interval level, we can use the formula for confidence interval calculation:

Confidence Interval = Mean ± (Z * (Standard Deviation / √Sample Size))

First, we need to find the value of Z for a 95% confidence interval. We can refer to the standard normal distribution table or use a statistical software to find the corresponding Z-value, which is approximately 1.96.

Now, let's plug in the values into the formula:

Confidence Interval = 4.5 ± (1.96 * (0.75 / √150))

Calculating the value inside the parentheses first, we get:

(0.75 / √150) ≈ 0.061

Then, multiplying that by 1.96, we get:

1.96 * 0.061 ≈ 0.1196

Finally, we can calculate the confidence interval:

Confidence Interval ≈ 4.5 ± 0.1196

This means that we estimate the mean range of all movies shown on television is between approximately 4.38 and 4.62 acts of physical violence at a 95% confidence interval level.