Question

PLEASE WRITE CLEARLY
A researcher reports survey results by stating that the standard error of the mean is 20. The population standard deviation is 500.

a.How large was the sample used in this survey?

b.What is the probability that the point estimate was within + or - 25of the population mean?

Answer 1

Standarderror =   

Here is standard deviation and n is sample size

So n = ( / S.E)^2

n = (500/20)^2

n= 625

So sample size should be 625 .

B)

Z score = ( x- m) / standard error

So lower z score is = -25/20= -1.25

Upper zscor is = 25/20 =1.25

Now p(-1.25<Z<1.25) = p(Z<1.25)- p(Z<-1.25)

Using z table

P(Z<1.25) = .8944

P(Z<-1.25)=.1056

So p(-1.25<Z<1.25) = .7888

So probability that point eatimate is within -+ 25 of population mean is .7888