Question

The mean of the medical test is 72.21  with the standard deviation of 2.5. Find a  95%confidence interval for a random sample of 25 students with the following scores​

Answer 1

Answer and Explanation:

The mean score for this medical test is 72.21 with a standard distribution of 2.5. We can create a 95% confidence interval for 25 students by using the graphical capabilities of Geogebra.  

The calculations indicate that the lowest point of the 95% confidence interval is 67 and the highest point of the 95% confidence interval is 77.

A random sample of 25 students would fall within the range of 67 to 77 with 95% confidence.

Answer 2

71.23 to 73.19     or    72.21 ± 0.98

If you know a population's standard deviation, you may compute a confidence interval (CI) for the population's mean, or average. When the population standard deviation is known, the formula for a confidence interval (CI) for a population mean is x ± z* σ/√n, where x is the sample mean, σ is the population standard deviation, n is the sample size, and z* represents the appropriate z*-value from the standard normal distribution for your desired confidence level.

Thus, Given:

Mean = 72.21

Standard Deviation = 2.5

Confidence level = 95%

Sample size = n = 25 students

Thus, using the formula,

Confidence Interval = Mean ± z* $\frac{standard deivation}{root n}$

= 72.21 ± 1.96 $\frac{2.5}{5}$

= 72.21 ± 0.98