Selected answers to Midterm Practice Questions, 302.

Hi all: I can't post the answers that are drawings (like histograms, distributions), but here are answers to selected questions:

1A. Most common errors: failing to label axes or switching axes around. Cups of coffee should be on the x-axis, frequency counts on y-axis.

Distribution is positively skewed and unimodal.

2. For pictures of distributions see chapters 2 & 3.

3A. Jerry's z-score. Formula is Z= (X-M)/SD

(4.5-3)/4 = 1/5/4 = .375 Jerry's z-score is +.375

3B. Fatima's score on the stats anxiety question.

Formula is X = (Z)(SD) + M

(-2)(4) + 3 = -8 + 3 = -5 Fatima's score is -5

(must be a scale that has negative as well as positive values!)

4A. 0 to 1.0

4C. -2 to +2 [see Figure 6.4 p. 139]. Or you could look up in unit normal tables for more precise numbers, but I'm testing the 2%, 14%, 34% rounded numbers for normal curve here.

6a. Z-test: You have one known population (myuu and sigma known) and one unknown population. You want to know if the two populations have different means. You take a sample from the unknown population.

6b. single sample t: Situation similar to z-test situation (see above) EXCEPT that you don't know sigma for the "known" population, only myuu.

6c. independent samples t: You have two unknown populations and want to know if they have different means. You take a sample from each population.

Hypothesis testing, with the 5 steps:

*NOTE* Before launching into the test, I check to see which test is appropriate. We have one sample (the first born children) from an unknown population, and we have a known population (myuu and sigma both known) to compare to. Hence a Z-test is appropriate. On the midterm, the question will call for EITHER z or single-sample t. Don't need to worry about the "by hand" calculations for independent t for the midterm.

1. State the two hypotheses, and select the alpha level:

Selecting a nondirectional test, so:

Ha: Myuu-firstborn not equal to Myuu-generalpop

Ho: Myuu-firstborn equal to Myuu-generalpop

Note: I consider it plausible that firstborn children might have lower IQs, even though this is not what I expect. Hence nondirectional hypotheses are appropriate.

Alpha. Set at .05, a commonly used value in psychology. [.01 would be fine, too]

2. Find critical value(s) and shade in critical region(s) on the sampling distribution:

To match nondirectional hypotheses, need 2-tailed test, hence there will be a critical region in each tail..

The shape of the sampling distribution should be normal, because underlying population (IQ scores) is normally distributed. For alpha of .05, critical values are z-scores of +/- 1.96 (you might have memorized this by now-or look up in unit normal table, being careful to remember that you want .025 in each tail). Mark these regions in the two tails.

[For alpha of .01, critical values are z-scores of +/- 2.58]

3. Collect data & calculate your test statistic (Assume you have collected data and have a sample mean for 20 first born children = 105).

The formula for your test statistic is Z = (x-bar - myuu)/SE

So you need x-bar, myuu, and SE

Myuu for known population = 100.

The standard error of the mean (SE) = sigma/sqrt of n = 15 /sqrt of 20 = 15/4.472 = 3.35

X-bar = 105 (just given above).

Z = (105-100)/3.35 = 5/3.35 = 1.49

4.Make decision about H_0:

Now you need to see where the Z test statistic falls on the sampling distribution. Is it in a shaded region? Then you reject the null. Another way to ask this: Is 1.49 a MORE EXTREME value than your critical values of +/- 2.58? If so, you reject the null.

If it does NOT fall in a shaded region (visual interpretation) or is LESS EXTREME than your critical values of +/- 2.58, then you "fail to reject " the null.

Use Z = (x-bar - myuu)/SE

Z = (105-100)/3.58 = 5/3.58 = +1.4 (rounded)

5. What's your answer to the research question, in ordinary English (not statistical jargon) based on this test?

Didn't find a significant difference in IQ for first born children at the 95% confidence level. Results are inconclusive. Note: DO NOT say "there is no difference." We don't know that. We only know that we didn't FIND a statistically significant difference in our test.

Question 8 (oops, misnumbering here, no question 8):

Here are answers to the first two questions that would receive full credit. They should give you a feel for how much you need to explain. If in doubt, explain more rather than less.

Why we make the n-1 correction when estimating population variance:

We use n-1 instead of n in the formula for sample variance SS/n-1 to correct the tendency of sample data to have a systematically smaller variance than the population. When we calculate the sample mean, x-bar, we lose a "degree of freedom" and minimize the variance compared to the true amount the scores vary from myuu (which will usually differ from x-bar). By using n-1, the degrees of freedom, instead of n, sample variance becomes an unbiased estimate of the population variance sigma-squared.

[NOTE: Be sure you read the question carefully. Notice this is not asking about df in relation to t-test, for example, only in relation to the n-1 denominator for estimating population variance]

Relation between sampling distribution, population distribution, and the distribution of a sample.

The sampling distribution is the distribution of means of a particular size drawn from the population distribution. The distribution of a sample is the distribution of the data in the sample, which is a subset of the data in the population distribution. Although a population distribution can have any shape, the sampling distribution will become more and more normal the larger the sample size. Both the population and the sampling distribution have the same mean, myuu, while the standard deviation of the sampling distribution (the standard error) will be smaller than the standard deviation for the population distribution.

NOTE: It's very important you clarify what each distribution is a distribution OF.

P. 5: This was covered in your lab. Be sure you know how to summarize the information in words and also how to present the t-test results in APA format.

For this output, it would look like this: Males ( M = 72.04, SD = 2.55) were significantly taller than females (M=64.51, SD=3.07), t(70) = 10.209, p < .001.

A few things to note: I gave a positive t value instead of the negative value, since the ordering of female and male is arbitrary. This is what's typically done in the literature. [If you gave a negative value, I wouldn't mark it as wrong.] I also didn't note that this was a two-tailed test, since this is the default approach. If you use a one-tailed test, however, you should always note this when reporting results. I also reported p < .001. If you said p < .05 or p < .01, that would also be correct. People ordinarily want to give the reader a sense of how strongly significant the difference was, but all are correct.