Question

You are interested in purchasing a new car. One of the many points you wish to consider is the resale value of the car after 5 years. Since you are particularly interested in a certain foreign​ sedan, you decide to estimate the resale value of this car with a 99​% confidence interval. You manage to obtain data on 17 recently resold​ 5-year-old foreign sedans of the same model. These 17 cars were resold at an average price of $ 12 comma 260 with a standard deviation of $ 800. Suppose that the interval is calculated to be (11693 comma 12827 ). How could the sample size and the confidence coefficient be altered in order to guarantee a decrease in the width of the​ interval?

Answer 1

If we want to guarantee a decrease in the width of the confidence interval we need to analyze the margin of error given by:

`ME = t_(\alpha/2) (s)/(√(n))`

The width of the confidence interval is just defined as two times the margin of error:

`Width = 2ME`

And then we can decrease this width of the interval we need to increase the sample size in order to have a greater number in the denominator for the margin of error and that implies a reduction in this width for the confidence interval at the confidence level fixed of 99%

By the other hand if we increase the confidence level then the width would be larger and in the other case when we decrease the confidence level the width would be lower.

Explanation:

For this case w ehave the following info given from the problem

`n =17` represent the sample size for the cars selected

`\bar X = 12260` represent the average price for the cars sold

`s= 800` represent the standard deviation for the solds

And we have a 99% confidence interval for the true average price for the cars given:

`11693 \leq \mu \leq 12827`

And we know that the confidence interval for the true mean is given by this formula:

`\bar X \pm t_(\alpha/2) (s)/(√(n))`

If we want to guarantee a decrease in the width of the confidence interval we need to analyze the margin of error given by:

`ME = t_(\alpha/2) (s)/(√(n))`

The width of the confidence interval is just defined as two times the margin of error:

`Width = 2ME`

And then we can decrease this width of the interval we need to increase the sample size in order to have a greater number in the denominator for the margin of error and that implies a reduction in this width for the confidence interval at the confidence level fixed of 99%.

By the other hand if we increase the confidence level then the width would be larger and in the other case when we decrease the confidence level the width would be lower.