Struggling Old Lady Needs Help..sorry long post

L

lala

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Feb 25, 2011
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Hi,
I'd like to briefly introduce myself. Math (and all things number related) have been a huge struggle for me my entire life. Simple things like phone numbers and street address get so jumbled up in my mind (and the street signs and piece of paper on which they're written) that I'm lost more often than not. My difficulties with math (and the resulting phobia) goes all the way back to 2nd grade when I used to get physically ill every day during the math period. I am old enough that algebra was not a required subject in high school, so of course I never made the attempt, and got no further than general math where I struggled through the basics of addition, subtraction, multiplication and division.

Having said all that, I have recently returned to school after a 30 year absence. I'm doing fairly well in most of my studies with the exception of - you guessed it - the math classes. I'm sure I sounded like a complete idiot the first day of pre-al when I asked why there were letters in math problem. LOL. At any rate, with a great deal of time (6-8 hours a day) and hard work I have made it as far as Statistics. I apologize for rambling about issues not pertinent to the math problem, but I figured if you knew my back round you might better understand that I need information broken down to a very basic level. So here we go...

This is the problem I'm currently struggling with:

"An experiment was conducted to compare the efficacies of two drugs in the prevention of tapeworms in the stomachs of a new breed of sheep. Samples of size 5 and 8 from each breed were given the drug and the two sample means were 28.6 and 40.0 worms/sheep. From previous studies, it is known that the variances in the two groups are 198 and 232, respectively, and that the number of worms in the stomachs has an approximate normal distribution. A 95% confidence interval for the the difference in the mean number of worms per sheep is?"

I know that the answer is supposed to be 11.4 +/- 16.2 but I have no idea how it was found. We are using a calculator for almost all of our work at this point and I've spent days punching numbers and have yet to find the correct sequence. I'm beginning to think there's a wicked 'trick' to this question that I'm not seeing. I was so sure that I should use 2-SampTInterval. I used n1 = 5, xbar1 = 28.6, Sx1 = 198, n2 = 8, xbar2 = 40, Sx2 = 232. Today I've sat and gone over the wording and now wonder if I'm using the wrong numbers. The problem says that we are comparing 2 different drugs on a new breed of sheep, but then it says we have samples from each breed. So are we comparing the 2 different drugs or are we comparing 2 different breeds? And if the sample sizes are only 5 and 8, how can the means be 28.6 and 40? I'm definitely not reading something right.

And as an aside (but probably equally important), in class when when worked these problems our answers didn't finish this way. All our instructor had us write was (this number) < mu1 - mu2 < (this number). That was the final answer. I assume the + / - 16.2 is the margin of error, but where does the 11.4 come from?

Again I apologize for rambling. I guess I just wanted to be clear that I'm not just looking for an easy answer to my homework...i am truly putting in the time and still struggling mightily.

Thanks,
La
 
The question is ambiguous - but it talks about "a new breed" and "two drugs". With that assumption 2-sample-T-test is correct.

What software are you using to calculate this?

Notice that the algebraic difference between those means is 11.4 (= 40 - 28.6).

Since difference encompasses 0, you cannot say - at the 95% significance level - that the efficacy of the drugs are different.
 
These are small sample sizes, 5 and 8. The means of 28.6 and 40 are the average number of worms per sheep from the samples taken.

If the sampling distribution for \(\displaystyle x_{1}-x_{2}\) is a t, and the population variances are not equal, we can use:

\(\displaystyle (x_{1}-x_{2})-t\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}<{\mu}_{1}-{\mu}_{2}<(x_{1}-x_{2})+t\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{1}}{n_{2}}}\) to find the confidence interval.

The d.f is the smaller of \(\displaystyle n_{1}-1\) or \(\displaystyle n_{2}-1\). Looking these up in the t table, sample size of 5, the t is 2.776 and sample size of 8, the t is 2.365.

I ran your sample sizes using a t of 2.365. This is the smallest t using the two sample sizes (with a d.f. of 7).

To get the confidence interval you state is correct \(\displaystyle -4.8<{\mu}_{1}-{\mu}_{2}<27.6\), the t should be 1.96. This is from a z distribution with a sample size >30.

Using the given sample sizes, I get a confidence interval of \(\displaystyle -30.1<{\mu}_{1}-{\mu}_{2}<7.3\)

Which goes from negative to positive. 0 is in the interval.
 
Thank you both SO much for responding. (That 11.4 was so simple I'm having a 'duh' moment.)

As for the rest of it....gibberish to me at this moment, but I have my text book by my side and have located the section that talks about the d.f. ( I SWEAR that was never covered in class.) But even if I decipher all the information, we are still at an impass. Using those figures ... -30.1 and 7.3 ....I get a margin of error of 18.7 (am I figuring that correctly?) I know the answer is 11.4 +/- 16.2 because I lucked out and guessed correctly on an online quiz and got it right. LOL. But should it be presented on an exam where I have to show my work I am out of luck.
 
OK! I've done some more piddling on the calculator (BTW I'm usuing a TI-84 plus) and discovered that If I find the square root of both those variences I can plug them into the 2SampInterval menu and get the same answer....7.28 and -30.08. That is probably why that df formula didn't look familiar. I'm guessing that my instructor never brought it up. He's all about technology and finding everything using the calculator.

At any rate I'm excited to be on the same page, but still bummed that our answer is supposedly not correct. I have a copy of my quiz, and my (guessed) answer of 11.4 +/- 16.2 is correct. Using the figures 7.28 and -30.08 our margin of error would be 18.68. Where are we picking up the additional 2.48?
 
As I stated, if the problem were a z distribution, then that answer would be correct.

Then we would have t=1.96. Which is the t score for a sample size of 30 or more.

At this level, the t and z scores are the same thing.

See, we use a z score when the sample size is large. Large being \(\displaystyle n\geq 30\).

This is why I am a little confused as well about the alleged correct answer and the small sample sizes.

If the problem stated that 50 were sampled from one group and 40 from the other for example, then OK.

But, as is, I am at a quandary as well. Those small sample sizes do not jive with that answer.

Unless, of course, we had a large sample. Then, we would be OK.

Look at the problem again and double check the sample sizes.
 
galactus I do so appreciate the time you've taken with me. Alas I have all the information correct (or at least as it was given to me) as I copy/pasted it directly from the quiz onto this thread.
 
I'm back! And I have the answer key with the problem in question (see attachment). Unfortunately I still don't see where those numbers are coming from. Any ideas?
 
As I stated earlier, those values come from a Z interval.

A large sample of 30 or more.
 
LOL OK. I was hoping that if you saw his figures you might see some logic with the answers, but apparently my instructor screwed up. Thanks again for your help.
 
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