A statistic question that's bugging me.

[Guest]

The claimed population mean of a country's household income is $1100.
I've sampled 100 people.
The sample mean income is $923.62
The sample standard deviation is $84.64

Population standard dev is not given.

So im proving or disproving the claim that $1100 is the true population mean at 1% significance level.

Ho: μ=1100
H1: μ≠1100

How do i go about doing this.

I used the T Test and got a weird answer of -20 something. Please help. thank you so much.

 

[Unco]

Hi AngelofDeath,

Form a 99% (guessing that that's what you mean by "1% significance level") confidence interval for the population mean based on your sample.

Use the confidence interval formula
pop mean = sample mean +/- Z * SD(pop)/(sqrt(n)).

Where
sample mean = 923.62;
we assume SD(pop) = SD(sample) so SD(pop) = 84.64
n = 100
Z = 2.576 (from the standard normal distribution tables)

If 1100 is included in this interval, then there is evidence to back-up the claim, otherwise there is evidence to suggest otherwise. You could consider how "far out" of the confidence interval the claimed mean is, if that is the case.

Hopefully this is stuff you are familiar with. I certainly don't know what the T test is. But upon googling, it looks like it's related to finding the difference of two population means, which is not what we are doing here.

 

[Guest]

Well, using the confidence interval, it only goes as high as 954.42 so 1100 is out of the question.
When I used the T test, it failed as well. I'm just not sure if i did it right. Thanks for your help Unco. If there are any other input, it's very welcom.

 

[tkhunny]

I don't see a critical value. Did you determine that? Is your design a one-tail or a two-tail test? That makes a difference. Why are you messing with a t-test. With a sample size as large as 100, skip it and move on to the Standard Normal. Unco demonstrated that.

Your test value is (1100-923.62)/84.64

Compare your test value to your critical value and determine which hypothesis to reject.

"-20" What are you doing? I'd really like to see how you managed that sort of thing. You didn't divide 84.64 by 10 again, did you?

 

[Guest]

If you look at my original post, I said I'm testing at 1% significance level.
It is a two tailed test.

I thought, you need to divide the standard deviation by the square root of n when testing. this is what I did.

Z test = (923.62-1100)/[(84.64/sqrt(100)]

I thought you have to divide the standard dev by n to get the standard error. That's what I had on my book.

I got -20. The z value for 1% significance level, two tailed test is -2.58. -20 is way off so I rejected the null hypothesis. What do you guys think?

 

[Unco]

As I said before, I can't help you with the T test, but I can say the standard error (which I'm assuming is universal in its definition here) is the standard deviation of the population divided by sqrt(n).

 

[tkhunny]

If you look at my original post, I said I'm testing at 1% significance level.

:roll:

It is a two tailed test.

At least you answered one question.

I thought, you need to divide the standard deviation by the square root of n when testing. this is what I did.

Good. Just checking, since 20 is such a large number. It just looks funny.

Z test = (923.62-1100)/[(84.64/sqrt(100)]

The value of the sample mean is, indeed, well below the expected population mean. Do you not like this ONLY because -20 looks so odd? I'd have to agree with your assessment. Personally, I'd go read the problem statement very carefully before I thought I was done.

 

[Guest]

Yeah. That was my problem. I'm stupified at -20. It seem to huge to be true. Usually in books, their's never go that far. Thanks for your help(s)