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A medical doctor wishes to test the claim that the standard deviation of the systolic blood pressure of deep sea divers is less than 450. To do so, she selected a random sample of 20 divers and found s = 432.
Assuming that the systolic blood pressures of deep sea divers are normally distributed, if the doctor wanted to test her research hypothesis at the .01 level of significance, what is the critical value?
Accepted Answer
To start with, standard deviations are NOT normally distributed. With that said, we'll continue.
I draw a line 3 feet long. How long is the line?
Well, obviously, 3 feet, right?
Wrong!! it is 36 inches. It is also 1 yard. It is also 1 / 1760 of a mile.
The point is that given some measure, there are usually MANY ways to describe that measure.
Now, take a distribution of some sort. There are also many ways to describe that distribution. We can "talk" about how many units are above a certain point, what percentage is above a certain point, or a "distance" along the horizontal axis (abscissa).
IF the "curve" is "normal", with some mean and standard deviation, there is a relationship between the percentages at different lengths along the abscissa.
Now, if you think about it, there are "many" normal curves with various means and standard deviations. It wouldn't be possible to relate those percentages to the various means/s.d. unless you could "standardize" those numbers. We thus "standardize" those numbers and call them "z scores", which we can then table. We standardize them by the following formula:
z = (score - mean) / s.d.
Let's look at a distribution with mean = 100, s.d. = 20. A score of 120 would have a "z score" of:
z = (120 - 100) / 20 = 20 / 20 = 1.00
If another distribution had a mean of 50 and s.d. of 10, a score of 60 would have a z score of:
(60 - 50) /10 = = 10/10 = 1
In some sense, a score of "120" on one distribution is "equivalent" to a score of 60 on another.
You can think of this as having one line 72 inches and another 2 yards. We can “standardize” the length in feet and find that each one is 6 feet long.
Now, there is one number you must memorize. 1 z score is the 84th percentile, which is 34% above the mean which is 50%. This memorization will allow you to use any z table (normal curve table). You see, different tables report the percentages different ways. When using any table, go to the z score of "1" and look at the number. If it is "84", the table reports the percentages from the left tail; if it is "34", the table reports the percentage from the mean. Alternatively, you can look at the value “0” and see if it is 0 or 50.
Now, we know the curve is symmetrical; and we know the mean is the 50th percentile; so let's play with it. If 1 s.d. is 34%, then, it must be the 84th %ile, because we have to add the 50% in for the mean. If 1 s.d. is 84, then there must be 16% ABOVE 1 s.d. Going the "other way", -1 s.d. must be 34% from the mean, or 84% from 1 s.d to the "right edge", so it must be the 16th %ile.
You should "play" with some more values, flipping back and forth above the value, below the value, using the left side of the mean and the right side of the mean. Drawing pics of the situation helps.
As you go through this course (and the next one) you will find that there are MANY "standard deviations". Although they have different names and different formulas, they ARE JUST "standard deviations". Hence, they will be called "standard error", standard deviation of the mean, standard error of a score, standard error of sampling, standard error of prediction, etc. All are used exactly the same way in relation to the normal curve.
NOW, let's look at YOUR problem
population mean = 450
Standard Deviation= 432
n = 20
alpha .01 (one tailed); z = 2.33
we now go to our basic formula:
(score - mean) / sd = 2.33
what is the sd of the mean? it is sd(pop)/sqrt(n)
(score - 432) / [432/sqrt(20)] = 2.33
(score - 432) / 96.6 = 2.33
score - 432 = 225.3
score = 432 ± 225.3 for a two tailed test. However, since you only wish the lower bound, use
432 - 225.3
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