Question

Assume that I asked heights of 40 female students. It turns out that average height and standard deviation of ladies in our class is 164 cm and 9 cm, respectively. For all computations, assume that the heights follow a normal distribution. For hypothesis test questions please use 5 step procedure.

a. (CI) Find a 95% confidence interval for the height of the female students.

b. (CI) Without making any computation, can you test whether ?=163 or not at ?=0.05? Why? If yes, what is the result and why?

c. (PI) Consider a lady that enters from the door. I say that her height is in the interval [a, b] with 95%. What is a and b? What kind of an interval is this?

d. (TL) Find the 99% tolerance limits that will contain 95% of ladies in this university.

e. (HT-Two Sided) Please test whether ?=165 or not by using a significance level of ?=0.01

f. (HT-Two Sided) For the question above, please plot the test statistic’s distribution function and show the observed value of the RV on the plot like we did in the lecture. You can plot it manually, no need for R.

g. (HT-Two Sided) Please test whether ?=164 or not using ?=0.01 level of significance.   

Can you please solve?

Answer 1

A) sample size = 40

sample mean = 164cm

standard deviation = 9cm

CONFIDENCE INTERVAL for population mean = sample mean +/- margin of error

margin of error = z_alpha/2 X SD/root(n) , (SD= standard deviation , n = sample size )

for 95% confidence interval , alpha % = (100-95)% = 5%

alpha/2 %=2.5% = 0.025

Z_0.025 = 1.95996

margin of error = 1.95996 x 9/root(40) = 2.789

CI = 164+/- 2.789 =    161.211 <height of girls lie between < 166.789

b) yes we can say that population mean = 163 at 5 % level of significance as it is conatined in the 95 % confidence interval that is we are 95% confident that population mean can be 163 , that is probability of population mean =163 is 0.95

c) this is a 95% confidence interval because you are 95% sure that her height will be in the range of values  between a and b

d) formula for tolerance limits

xbar+/- k x sd ( where x bar is sample mean , k is tolerance factor and sd is standard deviation)

for tolerance factor k we have following table

we need 99% tolerance limits with 95% coverage

so from the above table we have P=0.95 (COVERAGE ) ,confidence level = 99% and sample size=40

k = 2.684

Tolerance limits = 164+/- 2.684 x 9 = 164+/- 24.156 ,  [139.844 , 188.56 ]= tolerance interval