One-Sample z-test: Chapter 8

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One-Sample z-test: Chapter 8
Step-by-step procedure for the one-sample z test

We’re interested in testing the proposition – "IQ scores for SUNYA students are no different than the national average"

1. We state our formal hypotheses:

H₀: = 114 - saying that the UA mean IQ test score (for ALL UA students) will be 114 (so SUNY population = national population)

H₁: 114 - saying that the UA mean IQ test score is not equal to 114

- We want to determine which of these two competing hypotheses is more likely to be true.

2. We then get a sample (N = 150) of SUNYA students and measure their IQ scores – M = 117

- If we compute () = (117-114) = 3.00, we see that our sample mean differs from our population mean

- How likely is it that the difference we see is due to sampling error versus a "true" difference between UA students and the average college student?

This is where the one-sample z test comes in

- We’re interested in determining how likely it is that the mean intelligence level of SUNYA students is the same (given some sampling error) as the population mean for college students in general

- What is the probability that our sample mean is in the null sampling distribution of the mean vs. the probability that it belongs to some other sampling distribution of the mean

- The one-sample z test helps us make this judgment

*** Note: the following section is different than it is presented in the book.***

3. Before actually doing the statistical test, we need to develop our decision rules

- We set up rules beforehand to help decide when we will reject or not reject the null hypothesis

- Most often, our "critical values" (z_crit) are –1.96 and +1.96

A formal set of decision rules looks like this:

If z_obs > +1.96 or z_obs < -1.96, then reject H₀ (null hypothesis)

If –1.96 z_obs 1.96, do not reject H₀

Now that we have our decision rules, need to focus on calculating our z-value

- We use the following formula:

Basically, we are comparing the difference between the means to the average amount of sampling error- or the SD of the sampling distribution – or, the standard error of the mean.

5. Now we need to compute the standard error of the mean for the sampling distribution that corresponds to your sample size:

- Using the formula:

This tells us that, on average, we can expect any sample mean of N = 150 to deviate 1.225 units from the true population mean.

- We know that our sample mean deviates from the true population mean more than the average amount, BUT we don’t know if it is a large enough difference to say that SUNYA students are from a different population than the average student.

6. Need to convert our observed sample mean into a z-value to determine how many standard error units it falls from -- interpret like standard deviation

7. We then compare this observed z (z_obs) to your critical values (z_crit) to make the decision to reject or fail to reject the null

- If the observed results fall beyond the critical values, reject the null hypothesis

- Otherwise, do not reject the null hypothesis

WHAT IS OUR DECISION??

- 2.45 > +1.96, so decision is to reject H₀

If you reject the null hypothesis, compare the sample mean with the value of

stated in the null hypothesis and make conclusions about original research question

- In our study, our observed mean was statistically significantly greater than the population mean we compared it to

- Conclude that UA students have higher IQ scores than the average student.

- If you don’t reject the null – conclude that any difference you observed between your sample mean and your population mean is only due to sampling error.