can someone help me correct this statistics problem?

N

natenatmom

New member
Joined
Dec 17, 2005
Messages
24
Here is the question and how I answered. The teacher says it's wrong and I need to redo it but I'm confused!

Q: A weight loss program advertises "Lose 40 lbs in 4 Months." A random sample of n=25 customers has mean=32 pounds lost and s=12. Test null hypothesis = pop. mean = 40 against the alternate hypothesis = pop. mean = <40 with a significance = .05.

My answer was:

t = 32-40 = -2.5
12/sq rt of 25


R: T< -.5199

Null hypothesis is not rejected.

Here's what the teacher wrote: The t statistic and the rejection region are wrong. For t statistic, you substituted wrong values in the formula. Check the s value and the rest. Calculate rejection region and then justify the conclusion.


I really can't figure out what I did. I'm looking at the formula in my book for a small sample. This is a small sample, isn't it?
 
Yes, generally a sample is considered small if n < 30.

I don't know how you got -2.5 from (32 - 40)/(12/sqrt(25)).

Nor how you determined the rejection region to be t < -0.5199, which would correspond to a reasonably larger significance level than 0.05.

And, then, with your numbers, Ho would be rejected.
 
I have no idea! I've just gotten myself so confused now. I corrected my math error and got -3.33 for t. So is the rejection region .0005 then?
 
Go to your t-tables, go down to 24 degrees of freedom, move across to \(\displaystyle \L\mbox{t_{0.050}}\). You should see -1.711.
 
I got somewhat confused by your hypothesis notation. Did you mean:

\(\displaystyle H_{0}=40\)

\(\displaystyle H_{a}\neq{40}\)?

Looking this two-tailed test up in the t-distribution table(24 d.f. and \(\displaystyle {\alpha}\)=.05) we get t<-2.064 and t>2.064.

We arrive at -3.33.....

This is inside the rejection region, therefore, you opt to reject the null hypothesis.
 
So if the alt. hypothesis is <40 instead of not equal to, does that make the rejection region less than or equal to -1.711? And therefore the null hypothesis is rejected?

Also, why did my teacher say to check my s value? I copied that directly from the problem? How is my s value wrong?
 
I was wondering about the null hypothesis stating = and the alternate hypothesis

stating <. If the null hyp. is 'equal to', then the alternate should be 'not equal'

If the alternate is <, you would think the null would be \(\displaystyle \geq\).

Since \(\displaystyle H_{a}:{\mu}<40\), the claim, that makes it a left-tail.

\(\displaystyle H_{0}:{\mu}\geq{40}\). I don't know, something just seems a little

wacky here.

The rejection region is \(\displaystyle t<{-1.711}\). We found t=-3.33...

It is in the rejection region and so you reject the null hypothesis.

There is enough evidence at the 5% level of significance to reject the claim that you can lose 40 lbs.

As an extra check, your p value is .0014, which is less than your level of significance. When \(\displaystyle p<{\alpha}\) you reject \(\displaystyle H_{0}\)
 
Well this isn't the first time you all have discovered something wacky with this class and teacher! The question says exactly to "Test Ho = 40 against H1 < 40."
 
Since the statement is "lose 40 lbs in 3 months", I would think that means lose UP

TO 40 lbs in 3 months. So the null hypothesis would be \(\displaystyle H_{0}:{\mu}\leq{40}\) and the alternate hypothesis would be \(\displaystyle H_{a}:{\mu}>40\).

Or even, \(\displaystyle H_{0}:{\mu}=40\) and \(\displaystyle H_{a}:{\mu}\neq{40}\)

As a matter of fact, I recently conducted a statistical analysis at work to evaluate

the error in hi-tech balances. I used 'equal' and 'not equal' hypotheses.

I've never seen an 'equal to' null and a '<' alternate. Maybe someone else will

give their ideas on the matter. Unco?.
 
I'm inclined to go with the instructions. Ho: mu = 40, Ha: mu < 40.

I agree, Galact, that the Ho could just as well be mu >= 40. I think, though, we reason that if a mean of 40 can be rejected, then surely a mean greater than 40 can be rejected. So having an inequality for Ha and an equality for Ho is not uncommon.
 
OK Unco. I thought about that, but I had never seen them set up that way before.

Learn something new everyday.
 
Trig functions on the calculator

Help! This is an easy problem. But I can't use a calculator.


Solve the equation

secX=4.2 for X between 0 and 2pi (Radians)

I know I have to isolate the X. So the new equation would be

X=sec^(-1) (4.2)

I know I have to find the reference angle by putting it in the calculator. I don't know how to find the inverses of csc/sec/tan.

Please help. I've tried putting 1/cos then cos^(-1)(Ans) and vice versa. Either I'm punching it wrong or my whole method is wrong. :(

Thanks


P.S.
I don't think this has anything to do with it but I use a TI-84. :)
 
Trig functions on the calculator

Help! This is an easy problem. But I can't use a calculator.


Solve the equation

secX=4.2 for X between 0 and 2pi (Radians)

I know I have to isolate the X. So the new equation would be

X=sec^(-1) (4.2)

I know I have to find the reference angle by putting it in the calculator. I don't know how to find the inverses of csc/sec/tan.

Please help. I've tried putting 1/cos then cos^(-1)(Ans) and vice versa. Either I'm punching it in wrong or my whole method is wrong. :(

Thanks


P.S.
I don't think this has anything to do with it but I use a TI-84. :)
 
You've randomly posted in an existing thread.

sec(x) = 1/cos(x) so sec(x) = 4.2 --> cos(x) = 1/(4.2)

So cos<sup>-1</sup>(1/(4.2)) should work.
 
You must log in or register to reply here.
Top