1. A random sample of size 49 is taken from a large population, measuring the time it takes to complete a driver’s license examination. The sample mean was found to be 47 minutes, and the sample standard deviation 5.89 minutes. Construct a 95% confidence interval around the sample mean. Be sure to show whether you use the t- or z- value, and what value you use. Even though you are normally “allowed” to use an approximate value of “t” or “z” for this percent confidence, I want you to use the exact value appropriate for these circumstances. Interpret the confidence interval in a single sentence. In another sentence, state why you used either the “t” or the “z”.

Question

1. A random sample of size 49 is taken from a large population, measuring the time it takes to complete a driver’s license examination. The sample mean was found to be 47 minutes, and the sample standard deviation 5.89 minutes. Construct a 95% confidence interval around the sample mean. Be sure to show whether you use the t- or z- value, and what value you use. Even though you are normally “allowed” to use an approximate value of “t” or “z” for this percent confidence, I want you to use the exact value appropriate for these circumstances. Interpret the confidence interval in a single sentence. In another sentence, state why you used either the “t” or the “z”.

Answer 1

Check this answer.

http://www.jiskha.com/display.cgi?id=1279306420

Answer 2

can u be more clear on this

Answer 3

To construct a 95% confidence interval around the sample mean, we can use the formula:

Confidence interval = sample mean ± (critical value × standard error)

First, we need to determine whether to use the t-value or z-value. Since the sample size is relatively large (n = 49) and the population standard deviation is unknown, we will use the t-value.

To find the critical value, we need to determine the degrees of freedom, which is calculated as (sample size - 1). In this case, it is 49 - 1 = 48.

Using the t-distribution table or a statistical calculator, the critical value for a 95% confidence interval with 48 degrees of freedom is approximately 2.01.

Next, we need to calculate the standard error, which is the standard deviation of the sample divided by the square root of the sample size. In this case, it is 5.89 / sqrt(49) = 0.841.

Now we can calculate the confidence interval:

Confidence interval = 47 ± (2.01 × 0.841)
= (45.281, 48.719)

Interpreting the confidence interval, we can say with 95% confidence that the true population mean of driver's license examination time falls between 45.281 and 48.719 minutes.

We used the t-value in this case because the population standard deviation is unknown and the sample size is relatively small compared to the population size.