Question 1137293: A random sample is selected from a normal population with a mean of u=40 and a standard deviation of o=10. After a treatment is administered to the individuals in the sample, the sample mean is found to be M=46.
How large a sample is necessary for this sample mean to statistically significant? Assume a two-tailed test with a=.05.
If the sample mean were M=43 what sample size is needed to be significant for a two-tailed test with a=.05?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
I'll do the first part to get you started.
Population mean = mu = 40
Population standard deviation = sigma = 10
Because we know the value of the population standard deviation, we can use a normal Z distribution.
The formula for the z test statistic is
z = (M-mu)/(sigma/sqrt(n))
where n is the sample size and M is the sample mean (in this case M = 46).
Plug in the values mentioned and simplify
z = (M-mu)/(sigma/sqrt(n))
z = (46-40)/(10/sqrt(n))
z = 6/(10/sqrt(n))
z = (6/1)/(10/sqrt(n))
z = (6/1)*(sqrt(n)/10)
z = (6*sqrt(n))/10
z = (6/10)*sqrt(n)
z = (3/5)*sqrt(n)
Use a table or a calculator to find that z = 1.96 is the approximate critical value that corresponds to a 95% confidence interval, or put another way, you have 5% of the area in the two tails combined (2.5% in each tail). You can use this calculator if you don't have a TI calculator.
If you end up using the calculator I posted in the link, then follow these steps
- Type 0.05 for the area
- Enter 0 for the mean
- Enter 1 for the standard deviation (SD)
- click on the "outside" radio button
- click the "recalculate" button
The result "-1.96 and 1.96 will show up where you clicked on the "outside" ratio button. This means that P(Z < -1.96 or Z > 1.96) = 0.05 which is the combined area in the two tails. This combined area in the two tails is the alpha significance level we want.
We'll only focus on the positive z value. This is because M = 46 is larger than mu = 40, so the difference M-mu is positive, meaning the z score is above 0. Any z score larger than 1.96 will be considered statistically significant.
Plug z = 1.96 into the equation we found earlier. Isolate n
z = (3/5)*sqrt(n)
1.96 = (3/5)*sqrt(n)
1.96*5 = 3*sqrt(n) ..... multiply both sides by 5
9.8 = 3*sqrt(n)
3*sqrt(n) = 9.8
sqrt(n) = 9.8/3 ....... divide both sides by 3
sqrt(n) = 3.2667
n = (3.2667)^2 ...... square both sides
n = 10.6713
Keep in mind that the sample size n can only be a whole number.
Let's see what happens when n = 10
z = (3/5)*sqrt(n)
z = (3/5)*sqrt(10) ..... replace n with 10
z = 1.89736659610102
this value is not larger than z = 1.96, so a sample size of n = 10 tells us that a sample mean of M = 46 is not statistically significant.
Now plug in n = 11
z = (3/5)*sqrt(n)
z = (3/5)*sqrt(11)
z = 1.98997487421323
this value is larger than z = 1.96
So any sample size of n = 11 or larger will produce a z score that is larger than z = 1.96, meaning that we have a statistically significant result.
This is just for the sample mean M = 46. The steps for M = 43 will be similar.
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