Alcohol in Children’s Movies Listed below is a simple random s...

Question

Alcohol in Children’s Movies Listed below is a simple random sample of times (seconds) that animated children’s movies showed the use of alcohol (based on Data Set 20 “Alcohol and Tobacco in Movies” in Appendix B).

b. Are the requirements for constructing a 95% confidence interval estimate of the population standard deviation satisfied? If so, construct that confidence interval.

Accepted Answer

A media researcher selects a simple random sample of 15 PG-13 action films and records in seconds the duration of their longest explosion scene. The times are 47568, 10, 56765948, and 7. Assuming these durations are approximately normally distributed, are the conditions for constructing a 95% confidence interval for the population standard deviation sigma satis side. If so, compute that interval. Now, to solve this, we'll check our conditions first. Now to check our conditions. We first want to see if this is a random sample. And in this case, it tells us it is a simple random sample. So we can confirm We also need to check if we have a normal population. Now we can't check this directly, but we can. Assume if it is or not. And it does tell us that the durations are approximately normally distributed, so we can confirm this assumption. So since we confirmed both of those conditions, we can go ahead and find our confidence interval. Now, we have to find our mean and standard deviation first. Now this table here does have our values. But to find those are mean. We'll just be the sum of all of our values. Divided by the sample size, which in our case is 6. Or 7 Our standard deviation Will be the square root of the sum of each observation. Minus the mean Divided by sample size minus 1. And this difference in the numerator is squared. So for example, we will have the square root. Of our first observation Or Minus 6.47. All squared Plus 7 minus 6.47 squared, and so on until you get to the last value. Divided by 15 minus 1. Now the tip I provided calculated it for us, and that value is 1.767. So now we have mean and standard deviation. Next, we'll have to use our chi squared values. So, Our degrees of freedom Where chi squared in this case. Is our sample size minus 1, which is 14. And we can use a chi square table to find our specific critical values. In this case, Alpha equals 0.05 because we have a 95% confidence interval. This means we want the critical value. At alpha divided by 2. Degrees of freedom And also, the critical value at 1 minus alpha divided by 2 degrees of freedom. In this case, we'll have the chi squared. At 0.025 with degrees of freedom of 14. And the value. As 0.975 with degrees of freedom of also 14. And so now our confidence interval. will be In between Our two values. The square root of N minus 1 multiplied by variants divided by our lower chi square critical value. And the square root. Of minus 1 multiplied by variants divided by or upper. I squared value. Now, let's plug in our numbers. For our lower one we have the square root. A 14 Multiplied By standard deviation squared. Which is 1.767 squared. Divided by our critical value. Which from the table, this is 5.629. We'll do the same for the upper. Being 14, multiplied by 1.767 squared. Divided by our upper. Being 26.119. We can now calculate these with a calculator. We'll end up getting A value Of 1.294 for our lower bounds. And a value Of 2.787 for our upper bounds. Which will round to 1.29 seconds. And 2.79 seconds giving us our confidence interval. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Key Concepts

Confidence Interval

Standard Deviation

Simple Random Sample

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