Population means

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thordogg23

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Help, been working on this problem for the last week and can't seem to end up with a solid answer. any help would be greatly appreciated.

a questionaire was designed to compare the level of students familiarity with two types of products. For a random sample of 120 students, the mean familiarity level with burglar alarms was found to be 3.355 and the sample std. deviation was 2.03. In an independent random sample of 100 students, the mean familiarity level for television was 9.5, and the sample standard deviation was 2.1. Assuming that the two population distributions are normal and have the same variance, test the null hypothesis that the population means are equal
 
\(\displaystyle H_{0}:{\mu}_{1}={\mu}_{2}\)

\(\displaystyle H_{a}:{\mu}_{1}\neq {\mu}_{2}\)

You didn't state an alpha level, so I used .05

Running the data, I get a test stat of -22.009, critical value of \(\displaystyle \pm 1.96\), p value is very low. Well below any alpha level

we use even if it's .001. For that matter, look at the test stat. It is well within the rejection region. Therefore, it looks like a reject

the null hypothesis.
 
I used an Excel stats program I have. Sorry. I am not about to use those monster formulas. That is what calculators and Excel is for.

But, the formulas can be found in any stats book. Look under "testing the difference between means, large sample"

Or Google it. There is lots out there.
 
Well, I will show you what the formulas are. Like I said, they are cumbersome. A TI-83 will do these for that matter.

Please do not tell me you have some anachronistic, yahoo of a teacher who insists on doing thses by hand. If so, tell them I said to put down their slide rule and come into the 21st century.

We have a two-tailed test for this one. \(\displaystyle {\sigma}=\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}\)

Using the z-test to find the test stat:

\(\displaystyle z=\frac{(\overline{x}_{1}-\overline{x}_{2})-({\mu}_{1}-{\mu}_{2})}{{\sigma}}\)

But, \(\displaystyle {\mu}_{1}-{\mu}_{2}=0\) because \(\displaystyle H_{0}\) states \(\displaystyle {\mu_{1}={\mu}_{2}\)

Now, plug in your info and see if you get close to what I gave you.
 
For the first formula I came up with .23?

And for the second one -26.72?

?(2.032/120) + (2.12/100) = .23 standard error of difference in means
Z value = (3.355 – 9.5) –(0)/.23 = -26.72

Not sure why these are different than what you had. I am looking at my t distribution table and for .05 alpha and over 60 degrees of freedom I have 1.645? Is this correct? I see that you had 1.96.
 
Remember, I assumed .05. You did not state the alpha level. For .05 it is 1.96
 
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