Selected Answers to Problem Set 8

1. The alternate hypothesis should be regarded as very confident. Why?...
A level of confidence of 95%, for example, has a z-score of 1.645. That is, if our calculated z-score is greater than 1.645, then we are 95% confident that our alternate hypothesis is correct. To find the z-score, we take the difference between the found average and the null hypothesis average divided by the quotient of the standard deviation and the square root of the number of trials (similar to problem set 7, problem 2c).

2.7 / 1.25 = 2.16

We can see that 2.16 > 1.645, so we are 95% confident that the alternate hypothesis is actually correct.

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2. For this problem we will use a chi-squared test with double classifications.

Results

	Right Arm	Left Arm	Total
Right Thumb	52	79	131
Left Thumb	84	106	190
Total	136	195	321

From this we can calculate the expected values and decide on whether our null hypothesis (no correlation) is correct. To calculate the expected value of a cell we take the total for that hand times the total for that thumb, divided by the total number of people. For example, to find the expected value of right arm/right thumb, we take the number of right arm people (136) times the number of right thumb people (131) divided by the total number of people (321) to get (136 * 131 / 321) 55.5 people.

Calculations

		Right Arm	Left Arm	Row Totals
Right Thumb	Actual Values (A)	52	79	131
	Expected Values (E)	55.5	75.5
Left Thumb	Actual Values (A)	84	106	190
	Expected Values (E)	80.5	109.5
	Column Totals	136	195	321

Now we can find X². We now only have one degree of freedom, so we cannot really use the chi-squared test. We can correct for this by subtracting 0.5 from the difference between the actual value and the expected value. This correction may not have been presented in the notes; if not, don't worry about it, but have it in the back of your mind.

		Right Arm	Left Arm	Row Totals
Right Thumb	Actual Values (A)	52	79	131
	Expected Values (E)	55.5	75.5
	(|A-E| - 0.5)2 / E	0.162	0.0891
Left Thumb	Actual Values (A)	84	106	190
	Expected Values (E)	80.5	109.5
	(|A-E| - 0.5)2 / E	0.112	0.0822
	Column Totals	136	195	321

From the data we can see that X² = 0.445. We can then look up on a chi-squared chart for the normal score; in this case, with one degree of freedom and, say, 95% assurance, the score would be greater than 3.84. However, since it is less, we can disprove the alternate hypothesis that there is a correlation in the data. That is, we should assume that there is no correlation in this data. The second problem works out the same way.

		Right Arm	Left Arm	Row Totals
Men	Actual Values (A)	54	94	148
	Expected Values (E)	62.7	85.3
	(|A-E| - 0.5)2 / E	1.07	0.788
Women	Actual Values (A)	82	91	173
	Expected Values (E)	73.3	99.7
	(|A-E| - 0.5)2 / E	0.917	0.386
	Column Totals	136	185	321

Under this, X² = 3.158 and there is no correlation. However, without the -0.5 correlation, X² = 3.89, which is just above the 95% portion. With this one could find a correlation and perhaps find other ways of looking at the data.

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3. You can use my HTML script and you can read my Perl script here.