Any help would be greatly appreciated. I am one of the many students in my class that's lost, but we do what we can to help each other. Our instructor, although she seems very nice, isn't very helpful, or maybe we just can't grsped her way of teaching. Either way I am in dire need of assistance. Thank you in advance for your help.

Test the hypothesis that the mean temperature of women is 98.6 degrees. What can you conclude at the level of significance of a = 0.01?

I determined the mean to be 98.39; the standard deviation to be .7434877527; the margin of error to be .0922183061. I don't know where to go from here. Please help!


Testing Claims Using P-values
(a) write the claim mathematically and identify H0 and Ha.

I determined that Ha: u(less than or equal to)230 and Ho: u(greater than or equal to)230.
I also determined that it is one-tailed. What do i do next? I am completely lost..:confused:
(b) find the standardized test statistic z and its corresponding area.
(c) find the P-value.
(d) decide whether to reject or fail to reject the null hypothesis.
(e) interpret the decision in the context of the original claim.

In Illinois, a random sample of 85 eighth grade students has a mean score of 265 with a standard deviation of 55 on a national mathematics assessment test. This test result prompts a state school administrator to declare that the mean score for the state’s eighth graders on the examination is more than 260. At α = 0.04, is there enough evidence to support the administrator’s claim?

It would be a lot easier to do if you'd provide the information. For the women's temperature, you've determined 3 numbers that I have no clue where they've come from. You need a sample to test a hypothesis. But before doing the same, the first thing you need to do is set up what you are to test for (that is, Ho and Ha).

As for the second one, I don't know where you got the 230 from, nor could I tell you whether what you've done is correct because you haven't provided the information where this came from.

It would also be easier if you'd start with one problem, learn how to do them, and then try to apply it to another problem, instead of trying to do three things at once.

So stick with one thing and go back to the one about the women's temperatures. Can you set up your statement first? You need to determine if it's a one- or two-tailed test, which will then allow you to find your z score at the level of significance.

You should have already had z scores prior to this.

Yep, insufficient information. Anyway, you might want to check these out first under the heading: Statistics - Univariate Inferential Tests

Statistics - CliffsNotes

[QUOTE=mindenigma;

In Illinois, a random sample of 85 eighth grade students has a mean score of 265 with a standard deviation of 55 on a national mathematics assessment test. This test result prompts a state school administrator to declare that the mean score for the state’s eighth graders on the examination is more than 260. At α = 0.04, is there enough evidence to support the administrator’s claim?[/QUOTE]

Anyway: your null hypothesis is that the mean score for the state's 8th graders is less than or equal to 160. The alternative hypothesis is that it is more than 260. The administrator's claim (what you want to prove) is always the alternative hypothesis.

If the sample is large, by the central limit theorem, the sample mean score follows an approx normal distribution with mean 260 (under the null hypothesis) and variance 55^2/16

Calculate your z value and look up the cutoff z value for a one-tailed test at α = 0.04. If your calculated z value exceeds the cutoff z value, then you reject the null hypothesis and conclude that the administrator's claim is valid.

Quote:

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If the sample is large, by the central limit theorem, the sample mean score follows an approx normal distribution with mean 260 (under the null hypothesis) and variance 55^2/16

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Where did you get that variance from?

oops sorry :-/

I meant 55^2 divided by the sample size 85.

Well, OP seems to have disappeared, but what s/he needs is the standard deviation of the sample mean. i.e. the square root of that variance. I'm therefore not sure of the point of squaring 55 just to turn around and take a square root of it:





Or:



Just curious why the variance to begin with? It isn't used.

You are rite. My mistake.