angel_eyez88

New Member
(Q) a researcher is interested in whether a new cold medication affects mental alertness. the researcher administers the new cold medication to 16 individuals and then assesses their mental alertness using a standardized test. scores for alertness are normally distributed with population mean = 50 and SD = 12. the mean alertness for 16 indv is 55.5

(a) the researcher computes a two-tailed z-test on the data at alpha = 0.05, and infers that the cold medication has no significant effect on mental alertness. with a larger sample, the researcher may have drawn a different inference. assuming the sample was perfectly representative,what is the exact minimum sample size ( whole number value) required for correctly rejecting the null hypothesis?

---this is how i attempted the question, but didn't get the right answer, i don't know what im doing wrong or which other approach i should use:

we know for 2-tailed at 0.05 the z-values = + 1.96 and -1.96

now we must determine the new SEM ( standard error of mean) in order to find the new n ( sample size),

therefore we know z = M - mean / SEM
1.96 = 55.5-50/SEM
SEM = 2.81

now plug this into SEM formula = SD / sqroot(n)
2.81 = 12 / sqroot(n)
n = 18 ( apprx)

????? plz help == all feedback and help will be greatly appreciated == i need to know what im doing wrong -

Mean Joe

TS Contributor
what is the exact minimum sample size ( whole number value) required for correctly rejecting the null hypothesis?

n = 18 ( apprx)
One note: following the calculations you did, I got n=18.287.
You must round up, to find the minimum sample size. Do not round off. So if your calculations/reasoning is correct, n=19 is needed to reject the null hypothesis.

Dragan

Super Moderator
[/B]
(Q) a researcher is interested in whether a new cold medication affects mental alertness. the researcher administers the new cold medication to 16 individuals and then assesses their mental alertness using a standardized test. scores for alertness are normally distributed with population mean = 50 and SD = 12. the mean alertness for 16 indv is 55.5

(a) the researcher computes a two-tailed z-test on the data at alpha = 0.05, and infers that the cold medication has no significant effect on mental alertness. with a larger sample, the researcher may have drawn a different inference. assuming the sample was perfectly representative,what is the exact minimum sample size ( whole number value) required for correctly rejecting the null hypothesis?

---this is how i attempted the question, but didn't get the right answer, i don't know what im doing wrong or which other approach i should use:

we know for 2-tailed at 0.05 the z-values = + 1.96 and -1.96

now we must determine the new SEM ( standard error of mean) in order to find the new n ( sample size),

therefore we know z = M - mean / SEM
1.96 = 55.5-50/SEM
SEM = 2.81

now plug this into SEM formula = SD / sqroot(n)
2.81 = 12 / sqroot(n)
n = 18 ( apprx)

????? plz help == all feedback and help will be greatly appreciated == i need to know what im doing wrong -

In my opinion, I think there is something wrong with the way this question is presented. Specifically, the pharse: what is the exact minimum sample size...required for correctly rejecting the null hypothesis?.

This implies that H[0] is false. Thus, what I need to know is what is the researchers desired level of power 0.80, 0.90, etc.?? The appropriate (or minimum) sample size (N) is going to be contingent on this (Power) level.

For example, we know that d = (55.5 - 50) / 12 .

And so, Delta = d*Sqrt[N=16] = 1.833333 where Delta is the non-centrality parameter.

From a power table with alpha=0.05 (for a two-tailed test) we have approximately power = 0.44 (for N=16). This implies that we would be rejecting H[0] approximately 44% of the time - the other 56% of the time we would be making a Type II error.

If, however, I want to reject H[0] 80% of the time (an often used criterion), then I would need

N = (Delta / d)^2 = (2.8 / d)^2 = 37.3 or 38 subjects. Where the value 2.8 is from the power table. Thus, with N=38 subjects we would be rejecting H[0] 80% of the time and the other 20 percent of the time we would be making a Type II error.