Exam 8: T Tests

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a. H_o: Sample mean = population mean
H_A: Sample mean does not equal population mean.
b. 100 - 1 = 99
c. NOTE: Because it is 2-tailed, you must have a plus-minus critical value. This means that if your calculated, or observed, t value is either greater than 2.000 or less than -2.000, your result will be statistically significant. Also note that I used 60 degrees of freedom in Appendix B rather than 120. Even though 99 is closer to 120 than it is to 60, go with the smaller number when you have a df between two numbers. If your results are significant at the smaller df, they will be significant at the larger df.
d. First I need to calculate a standard error of the mean:
s_x= 3/?100 $\rightarrow$ 3/10 $\rightarrow$ .30

t₀ = (6 - 5)/.30 $\rightarrow$ 1/.30 $\rightarrow$ 3.33.
e. I am going to REJECT the null hypothesis because my observed t value exceeds my critical t value. IN other words, 3.33>2.000, so reject the null hypothesis
f. Yes.
g. The chances of obtaining my results by chance were less than 5% (p < alpha), therefore I decide my results are statistically significant.
h. CI₉₅ = 6 $\rightarrow$ (.30) (2.000) $\rightarrow$ 6 $\rightarrow$ .60 $\rightarrow$ 5.40, 6.60. I am 95% confident that the average level of happiness in the population of Californians is between 5.40 and 6.60. Notice how this interval does not include the population mean of Americans, which was 5? We already knew this because our result was statistically significant, meaning that the sample mean was different from the population mean at the .05 level. Also notice that our population we are talking about in this confidence interval is the population of CALIFORNIANS. We are concluding that the population of Californians differs in happiness from the population of Americans. Notice how we used sample data to reach a conclusion about populations? That is the crux of inferential statistics.

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a. Independent samples t test.
b. 1-tailed, because the researcher clearly expects Michigan students to score higher than BU students, on average.
c. Michigan mean > BU mean.
d. If I make Michigan the X1 variable and BU the X2 variable, I would expect a positive t value (because I expect Michigan students to have the higher mean). If the Xs are reversed, I would expect a negative t value. Using an alpha level of .05, 1-tailed, and 120 df (61 + 61 -2 = 120), my critical t value would be 1.658.
e. First I need to calculate standard errors of the mean for each sample.
For Michigan,
$S_{\bar{x}}$
= 200/7.81 =25.61. For BU students,
$S_{\bar{x}}$
= 150/7.81 = 19.21.
Then I square each
$S_{\bar{x}}$
: (25.61)² = 655.87. (19.21)² = 369.02. When I add these together I get 1024.89. Take the square root of that and I get 32.01. So my observed t = 600 - 565 / 32.01 $\rightarrow$ 35 / 32.01 = 1.09. So my observed t value of 1.09 is less than the critical t value of 1.658. Therefore, I conclude that, in the population of Michigan students and BU students, the average SAT scores are equal. The difference in sample means is too likely due to chance for me to conclude that it was not.
f. Maybe Michigan undergrads really aren't any better than BU undergrads. Ugh.

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So I do not know the population standard deviation here, which means I'll have to look in Appendix B using the t values. First, I must calculate a standard error of the mean:
$S_{\bar{x}}$ = 4/? 16 $\rightarrow$ 4/4 $\rightarrow$ 1.
Next I calculate a t value: t = 10 - 12/1 = -2/1 = -2.00.
My degrees of freedom will be 16 = 1 = 15.
So in Appendix B, looking at 15 degrees of freedom I find a value of 1.753 and a value of 2.131. These are the two closest values to 2.00. Then I look up at the top of the table and find that these correspond with alpha levels of .10 and .05. So the best I can do is say that the probability of getting a sample mean of 10 from this population, when the sample is randomly selected and n = 16, is between 5% and 10%.