A confidence interval is found by adding and subtracting the margin of error.
If we take the difference of 235.1 and 197.2 and divide by 2 we have the margin of error.
this margin of error is found by taking 1.96 times the standard deviation.
To find the standard deviation divide 1.96 in to the answer from above. Now, you will have the sd for the next part of the problem.
A study of 29 female Sumatran elephants provided a 95% confidence interval for the mean shoulder height as (197.2, 235.1) cm. Consider a new study to estimate the mean shoulder height of male Sumatran elephants. Assuming the standard deviations of shoulder heights are similar for males and females, how many male elephants should be sampled so that the 95% confidence interval for the mean shoulder height of males will have a margin of error of 10 cm?
I know the formula to determine the sample size is:
n=((1.96∙?)/10)
How do I guess the Standard Deviation? I cannot find anything in any scientific literature that has a similar SD.
Thanks for you help.
4 answers
Thanks John :)
You are welcome.
"John"'s answer is incomplete. The real solution is:
W_female: (235.1 - 197.2)/2 = 18.45
t-value for females: 2.048
SD_female: (W*sqrt(n))/t = (18.45*sqrt(29))/2.048 = 48.5
As stated, the margin of error for males is 10. We assume the t-value for males is 2 based on the female value. So,
10 = 2*(48.5/sqrt(n))
n = 9409/100 = 94.09 male elephants
OR you can say:
n = (4.85 * 2)^2 = 94 male elephants.
Either way, this is your answer.
W_female: (235.1 - 197.2)/2 = 18.45
t-value for females: 2.048
SD_female: (W*sqrt(n))/t = (18.45*sqrt(29))/2.048 = 48.5
As stated, the margin of error for males is 10. We assume the t-value for males is 2 based on the female value. So,
10 = 2*(48.5/sqrt(n))
n = 9409/100 = 94.09 male elephants
OR you can say:
n = (4.85 * 2)^2 = 94 male elephants.
Either way, this is your answer.