Question 1199299
population mean is assumed to be 162.5 centimeters with a standard deviation of 6.9 centimeters.
sample of 50 females has an average of 165.2 centimeters.
level of significance is .05.
since the test is for not equal, the confidence interval has two tails, each tail is equal to .025.
the z-score formula is z = (x-m)/s
z is the z-score
x is the sample mean
m is the mean
s is the standard error which is the standard deviation of the distribution of sample means.
z-score formula becomes:
z = (165.2 - 162.5) / standard error which is equal to standard deviation / square root of sample size which is equal to 6.9/sqrt(50) = .975807358.
z-score is therefore equal to 2.767 rounded to 3 decimal places.
the area to the right of that z-score under the normal distribution curve is equal to .0028 rounded to 4 decimal places.
since .0028 is less than .025, the results are significant and the conclusion is that there is a very high probability that the average height of females in the freshman class is not equal to 162.5.


the normal distribution graph results are shown below.


<img src = "http://theo.x10hosting.com/2022/121902.jpg">