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”.

please put answer more clear on this

If your sample size is large (n is greater than or equal to 30), use z-value.

If you have a small sample (n < 30), then use t-value.

CI95 = mean + or - 1.96(sd divided by √n)
...where + or - 1.96 represents the 95% confidence interval using a z-table, sd = standard deviation, √ = square root, and n = sample size.

Plug the values from the problem into the formula to find your confidence interval.

I hope this will help get you started.

To construct a 95% confidence interval around the sample mean, we first need to determine whether to use the t-value or the z-value. Generally, the t-value is used when the population standard deviation is unknown (and estimated using the sample standard deviation), and the z-value is used when the population standard deviation is known. In this case, since we are given the sample standard deviation but not the population standard deviation, we'll use the t-value.

The formula to calculate the confidence interval for the mean is:

Confidence Interval = sample mean ± (t-value * standard error)

First, we calculate the standard error:

Standard Error = sample standard deviation / √(sample size)

Plugging in the given values, we have:

Standard Error = 5.89 / √(49)
≈ 0.839

Next, we need to find the t-value. Since the sample size is 49, which is relatively large, we'll use the t-distribution with 49 - 1 = 48 degrees of freedom. The critical value for a 95% confidence level and 48 degrees of freedom is approximately 2.010.

Now we can calculate the confidence interval:

Confidence Interval = 47 ± (2.010 * 0.839)
= 47 ± 1.687
= (45.313, 48.687)

Interpretation of the confidence interval: We are 95% confident that the true population mean for the time it takes to complete a driver's license examination lies between 45.313 minutes and 48.687 minutes.

We used the t-value because we do not know the population standard deviation and had to estimate it using the sample standard deviation.