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).

Question

Quizplus · Accepted Answer

The Answer is 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.

I Know That Men Are Usually the Gender Portrayed as All