Deck 8: T Tests

Question 1

I know that men are usually the gender portrayed as all amped up and hyper, but in my experience as a commuter (from Oakland to Santa Clara) I must say that it seems women drive faster than men, on average. To see if I am correct, I solicited the help of the Highway Patrol one day. I sat with an officer and we used a radar gun to measure the speeds of 16 randomly selected male drivers and 16 randomly selected female drivers on the same stretch of Interstate 880. We found that our sample of male drivers had an average speed of 68 mph with a standard deviation of 15. Our sample of female drivers had an average speed of 71 MPH with a standard deviation of 10.
b. State the null and alternative hypotheses
c. Chose an alpha level.
d. What exactly does this alpha level mean?
e. Decide whether you are doing a 1-tailed or 2-tailed test (explain why)
f. State your degrees of freedom
g. Find and report your critical value for t.
h. Compute your observed t-value.
i. Wrap words around your results. Is this a statistically significant difference?
j. Calculate an effect size (d).

Answer 1

b. Ho: $\mu$ ₁ = $\mu$ ₂
H₁: $\mu$ ₁ > $\mu$ ₂
Sample 1 is the female sample, Sample 2 the male
c. Either .05 or .01. I'll solve for both.
d. Percentage of time I'm willing to be wrong when I reject my null hypothesis and conclude that the results were not due to chance.
e. One-tailed because I am speculating that women will drive faster than men, on average.
f. df = (16 + 16 - 2) = 30 (use 30 from App. B)
g. 1.697 for alpha of .05, 2.457 for alpha of .01. Note that these are the critical t values for 1-tailed test, because my alternative hypothesis is directional. Note that I am only interested in positive t values for this test because if my alternative hypothesis is true, it will produce a positive t value.
h.I first need to calculate two standard errors:
For women,
$S_{\bar{x}}$
= 10/ $\surd$ 16 $\rightarrow$ 10/4 $\rightarrow$ 2.50
For men,
$S_{\bar{x}}$
= 15/ $\surd$ 16 $\rightarrow$ 15/4 $\rightarrow$ 3.75
Now, to find the standard error of the difference between the means I need to square each standard error, then add them together, then square root this sum:
(2.50)² = 6.25
(3.75)² = 14.06
6.25 + 14.06 = 20.31. $\surd$ 20.31 = 4.51
Now I can compute my observed t value: 71 - 68/4.51 $\rightarrow$ -3/4.51 $\rightarrow$ -.67.
i. No, it is not statistically significant. I fail to reject H_o because my observed t value does not exceed than my critical t value. The chances of finding a difference between the sample means of this size by chance is greater than my alpha level. Therefore I cannot conclude that there is a difference between the population means of men and women in the liking of statistics.
j. To calculate an effect size we need to know the two sample means, the standard error of the difference between the sample means, and the size of one of the samples:
68 = mean for men, 71 = mean for women, 4.51 = standard error, and n = 16.
The square root of n = 4. (4)(4.51) = 18.04
So d = (68 - 71)/18.04 $\rightarrow$ -3/18.04 = -.17. This is considered a pretty small effect size, so the results would not be considered to be of much practical significance.

Question 2

Ahh, the holidays-you gotta love 'em. Nothing says holiday quite like eating too much food. Whether it is Thanksgiving, Christmas, or the 4^th of July, we love to eat on holidays. I wonder whether people spend more money at the grocery store two weeks before a major holiday or two weeks after. So I select a random sample of 25 people and see how much they spend on a trip to the store two weeks before the holiday and again two weeks after. Before the holiday, the shoppers spent an average of $80. After the holiday, each shopper spent $85, on average. The standard DEVIATION of the DIFFERENCE between the means was 10.
a. Write the null and alternative hypotheses.
b. Select an alpha level and wrap words around it.
c. Is this a statistically significant difference? (do the calculations)

Answer 2

a. H₀: $\mu$ _x = $\mu$ _y
Note that we need to specify here that the X population is shoppers before the holidays and the Y population of shoppers after the holidays. Need to be clear on which population is which.
H₁: $\mu$ _x $\neq$ $\mu$ _y Note that I used this non-directional alternative hypothesis because the question says they could spend more before OR after the holidays
b. Alpha is .05, meaning that there must be less than a 5% chance that my results are due to chance, or random sampling error, before I will consider them statistically significant.
c. First I need to find the standard error of the differences between the sample means. I'm already given the standard deviation of the difference. So I take that number and divide it by the square root of N, which is the number of pairs in the sample. So:
$s_{\bar{D}}$ = 10 / $\surd$ 25 $\rightarrow$ 10/5 $\rightarrow$ 2
Now I calculate the observed t value:
$t=\frac{\bar{X}-\bar{Y}}{s_{\overline{D}}}$ $\rightarrow$ $t=\frac{80-85}{2}$
$\rightarrow$ $t=\frac{-5}{2}$ $\rightarrow$ t = -2.50. Because this is a 2-tailed test, I can take the absolute value and get an observed t value of 2.50.
Using Appendix B, 2-tailed test, .05 alpha, df = 25 - 1 = 24, I find a critical value of ±2.064.
Because my critical t value is less than my observed t value, I reject the null hypothesis and conclude that these results were not due to chance. In the populations of grocery shoppers, people spend more money on average after the holidays than they do before.

Question 3

Boys and girls have a lot in common, but they also seem to have a lot of differences. Some people have argued that, even at a young age, girls are more nurturing and caring than boys. Others think this is just an unfounded stereotype. Suppose that you, as a developmental psychologist, decide to test this out. You select a random sample of 20 boys and 20 girls and measure their level of caring for a sad friend. Boys have a mean of 4.0 and a standard deviation of 1.0 on the nurturing scale. Girls have a mean of 4.5 and a standard deviation of 1.2 on the nurturing scale. Using an alpha level of .05, conduct the analysis to determine which hypothesis is supported.

Answer 3

X_b = 4.0, s_b = 1.0
Xg = 4.5, s_g = 1.2
df = 20 + 20 -2 = 38
$\alpha$ = .05
2-tailed
t_c = 1.69
$\begin{array}{l}s_{\bar{x}-\mathrm{b}}=1 / \sqrt{20} \rightarrow 1 / 4.47 \rightarrow .22 \\s_{\bar{x}-\mathrm{g}}=1.2 / \sqrt{20} \rightarrow 1.2 / 4.47 \rightarrow .27 \\\\s_{\bar{x} 1-\bar{x} 2}=\sqrt{ }(.22)^{2}+(.27)^{2} \rightarrow \sqrt{ }(.048+.073) \rightarrow \sqrt{.121} \rightarrow .348 \\\\t_{o}=4.5-4.0 / .348 \\t_{o}=.5 / .348 \\t_{o}=1.44\end{array}$ My observed t is smaller than my critical t. Therefore, I retain the null hypothesis and conclude that there are no differences between boys and girls in their level of caring for a sad friend in the populations.

Question 4

Please write a research question that would be best answered with a 2-tailed, dependent-samples t test.

Question 5

Please write a research question that would most appropriately be answered with a dependent samples t test.

Question 6

In all of the different t tests, we use the same general formula: observed effects in the sample divided by a standard error. What does such a formula tell us and why does it make sense as a way to test whether a result is statistically significant?

Question 7

Suppose that in the population of American college students the average number of alcoholic drinks consumed per week is 12. What is the probability of randomly selecting a sample (n = 16) with a mean of 10 and a standard deviation of 4? (Hint: Tell me the two probabilities it will be between using a 2-tailed value)

Question 8

When I was a student at the University of Michigan, we had a little saying that we liked to use with students who attended other universities: "We're not snobs-we're just better than you." To prove it, I collected data from 61 randomly selected undergrads at Michigan and 61 randomly selected undergrads at Boston University (BU). The Michigan sample had a mean SAT verbal score of 600 with a standard deviation of 200. The BU sample had a mean SAT verbal score of 565 with a standard deviation of 150.
a. What kind of test should I use to examine whether the difference between these means is statistically significant?
b. Is this a 1-tailed or a 2-tailed test? Why do you think so?
c. What would your alternative hypothesis be? Tell me why.
d. Report the critical value for this test. Why do you think that is the correct critical value
e. So perform the necessary calculations and tell me whether the difference between the means is statistically significant using an alpha level of .05. Be sure to wrap words around your results. Tell me all that you know
f. Give me one plausible explanation for your results. Why do you think the results turned out this way. (Note: I am not looking for the technical, mathematical answer here-nothing about probabilities or formulas. I just want you to give me a reasonable explanation for why Michigan students may have higher SAT scores, or not, depending on what your results were.)

Question 9

California is a beautiful place to live, but it is also crowded and expensive. Given the good and the bad of living here, I wonder whether Californian's, on average, differ from other Americans in how happy they are. I happen to know that on a 10-point scale (1 = miserable, 10 = extremely happy) American adults have an average score of 5. I select a random sample of 100 Californian adults and find that they have an average score of 6 with a standard deviation of 3 on the happiness scale.
a. State the null and alternative hypotheses
b. State your degrees of freedom:
c. Find and report your critical value for t. For an alpha level of .05, t_c= $\pm$ 2.000.
d. Compute your observed t-value.
e. Decide whether to reject or fail to reject H_o .
f. Is this a statistically significant difference?
g. What does that mean, exactly?
h. Calculate and report a 95% confidence interval for the sample mean and wrap words around it."