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Confidence intervals for one population mean when S.D. is unknown.

A Variable has a mean of 100 and a standard deviation of 16. Four observations of this variable have a mean of 108 and a sample standard deviation of 12. Determine the observed value of the a. Standardized version of x (sample mean). b. Studentized version of x (sample mean).

Subject:

Statistics

Topic:

All Topics

Posting ID:

4509

OTA ID:

103060

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Hypothesis Testing: Compute the statistical z test and deciding what to do with the null.

In a recent national survey, the mean weekly allowance for a nine-year-old child from his/her parents was reported to be $3.65. A random sample of 45 nine-year-olds in northwestern Ohio revealed the mean allowance to be $3.69 with a standard deviation of 0.24. At the 0.05 level of significance is there a difference in the mean allowances nationally and the mean allowances in northwestern Ohio for nine-year-olds. Test = z test State the research hypothesis and the Null hypothesis.

Subject:

Statistics

Topic:

All Topics

Posting ID:

4510

OTA ID:

101733

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Hypothesis Testing: Compute the statistical test using the t test and decide what to do with the null.

One of the Rocky Mountain School District Principal's claims that each of his assistant principals makes 40 calls on parents per week. Several other Principals said that this estimate is too low and that their assistant principals make even more calls. To research the claim, a random sample of 28 assistants revealed that mean number of calls made last week was 42. The standard deviation of the sample was computed to be 21 calls. At the 0.05 level of significance can we conclude that more than 40 calls are made on the average in a week? State the research Hypothesis and the Null Hypothesis. Test = t test

Subject:

Statistics

Topic:

All Topics

Posting ID:

4512

OTA ID:

101733

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finding expected value in 2x2 table

Myocardial Infarction Status Yes No OC User 683 2537 Non-OC user 1498 8747 The question is what the expected value for an OC user and no myocardial infarction? I found all kinds of things to calculate in 2x2 tables, but not this. Probably am overlooking something simple.

Subject:

Statistics

Topic:

All Topics

Posting ID:

4513

OTA ID:

103139

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Question on expected values

Given the following 2x2 table: Myocardial Infarction Status Yes No OC User 683 2537 Non-OC user 1498 8747 Calculate the expected value for an OC user and no myocardial infarction. The TA said "2537", which may be correct, but I think some kind of chi-square goodness of fit or a similar test is needed to see what the "expected value" would be or how it would fit a normal distribution. I have waded through some reference material that would indicate there is a way to solve for the expected value, given the sample, but am unclear on how to work this.

Subject:

Statistics

Topic:

All Topics

Posting ID:

4551

OTA ID:

102922

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