Hyposthesis test

E

elquicko

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Joined
Nov 17, 2013
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Forgive my English, if its not too good.

I'm comparing two separate populations. Can it be concluded at a 5% significance level that the lung capacity of smokers is less than nonsmokers? There are 10 persons tested in each population.
I calculated the numbers (I won't list them all here)

For the Nonsmokers: We have a
Mean = 1.08 and SD = .1515

For Smokers: We have a
Mean = 1.004 and SD = .1319


I think this test is a one tailed test. H0: m = 1.08
HA: m < 1.08

So I took standard error of the mean: std dev / Sqrt of n

.1319/ 3.162 and standard error of mean is: .041

Then i did Test statistic: 1.004- 1.08 / .041 = -1.854

That gives me a Z-score of .0322

Since it falls into the critical region of .05 it is safe to say there is a difference and we reject our null hypothesis.

Can someone take a look at this please? Thank you very much.
 
elquicko said:
Forgive my English, if its not too good.

I'm comparing two separate populations. Can it be concluded at a 5% significance level that the lung capacity of smokers is less than nonsmokers? There are 10 persons tested in each population.
I calculated the numbers (I won't list them all here)

For the Nonsmokers: We have a
Mean = 1.08 and SD = .1515

For Smokers: We have a
Mean = 1.004 and SD = .1319


I think this test is a one tailed test. H0: m = 1.08
HA: m < 1.08

So I took standard error of the mean: std dev / Sqrt of n

.1319/ 3.162 and standard error of mean is: .041....why use sigma of just the smoking group??

Then i did Test statistic: 1.004- 1.08 / .041 = -1.854....= z-score

That gives me a Z-score of .0322....not the z-score, but rather the probability of that z is random;
........................................... ..........the critical value for 5% in the lower tail is z=-1.645

Since it falls into the critical region of .05 it is safe to say there is a difference and we reject our null hypothesis.

Can someone take a look at this please? Thank you very much.
Click to expand...
I do things a little differently, but the result should be similar. The main difference is what I use for the standard deviation for getting the z-score.

Under the null hypothesis, the difference of the two means is zero. Since the two data sets are independent, the Variance of the difference is the sum of the Variances of the two groups, giving the standard error of the mean difference = 0.064. That reduces the z-score to -1.2, which does not require rejection of the null hypothesis at the 5% confidence level.
 
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