9.73 Statistics on Large Sample Tests of Hypotheses?

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The discussion centers on testing whether the average SAT verbal and math scores for California students in the class of 2003 differ from national averages, using a sample of 100 students. For the verbal scores, the sample average is 499, while the national average is 507, leading to a calculated z-score of -0.816, which is incorrect according to the book's answer of -3.33. The correct calculation involves using the standard error, which is derived from the standard deviation divided by the square root of the sample size. The math scores show a sample average of 516, and the discussion also questions whether the data can determine differences between math and verbal scores. Ultimately, the participants seek clarification on the correct z-score calculation and its implications for hypothesis testing.
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Homework Statement


a. Do the data provide sufficient evidence to indicate that the average verbal score for all California students in the class of 2003 is different from the national average? Test using alpha = .05.
b. Do the data provide sufficient evidence to indicate that the average math score for all California students in the class of 2003 is different from the national average? Test using alpha = .05.
c. Could you use this data to determine if there is a difference between the average math and verbal scores for all California
students in the class of 2003? Explain your answer.

Homework Equations


Relevant DATA:
n = 100 (California students sampled)
national average scores for verbal = 507
sample California average for verbal = 499
standard dev California for verbal = 98

Relevant EQUATION:
z = [sample California average - mean National]
[standard dev Californial for verbal/sqrt(n)]

The Attempt at a Solution


z = [499 - 507]/[98(sqrt(100))]
= -8/9.8
= - 0.816

This answer for part a. is wrong.

Answers at the back of the book:
a. yes; z = -3.33 b. no; z = 1 c. no

I am stuck at part a. My wrong answer is z = -0.816. Please show me how step-by-step to derive at the correct answer. Thank you very much.
 
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I'm confused. You say "My wrong answer is z = -0.816" but none of these questions asks for a value of z! What have you done to answer a?

In any case, I will point out that 98\sqrt{100}= 980, not 9.8. The correct z-score is much less than -0.816. (That's assuming that your figure of 100 for the standard deviation is correct. It seems awfully high for average scores around 500. It would pretty much mean that the test is meaningless!)
 
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Little Bear said:

Homework Statement


(I forgot to write out the problem statement too!)
How do California H.S. students compare to students nationwide in their college readiness, as measured by their SAT scores? The national average scores for the class of 2003 were 507 on the verbal portion and 519 on the math portion. Suppose that 100 California students from the class of 2003 were randomly selected and their SAT scores recorded as:

Verbal: Sample average = 499, Sample standard dev = 98

Math: Sample average = 516, Sample standard dev = 96




a. Do the data provide sufficient evidence to indicate that the average verbal score for all California students in the class of 2003 different from the national average? Test using alpha = .05.is different from the national average? Test using alpha = .05.
b. Do the data provide sufficient evidence to indicate that the average math score for all California students in the class of 2003 is different from the national average? Test using alpha = .05.

c. Could you use this data to determine if there is a difference between the average math and verbal scores for all California
students in the class of 2003? Explain your answer.


Homework Equations


Relevant DATA:
n = 100 (California students sampled)
national average scores for verbal = 507
sample California average for verbal = 499
standard dev California for verbal = 98

Relevant EQUATION:

SE = standard dev/sqrt(n)
z = [sample California average - mean National]
[standard dev Californial for verbal/sqrt(n)]
OR
z = [sample California average - mean National]/ SE




The Attempt at a Solution


z = [499 - 507]/[98(sqrt(100))]
I forgot the the "/" between 98 and sqrt(100)
For the denominator of z is actually standard error = standard dev/n
Since, standard dev of california sample for verbal test = 98 and n = 100 samples
SE = standard dev/sqrt(n) = 98/sqrt(100) = 9.8
Thus z = [mean of sample California - mean of National (Popn)]/SE
= [499-507]/9.8
= -8/9.8
= - 0.816

This answer for part a. is wrong.

Answers at the back of the book:
a. yes; z = -3.33 b. no; z = 1 c. no
I am stuck at part a. My wrong answer is z = -0.816. The back of the book says -3.33. Please show me how step-by-step to derive at the correct answer. Thank you very much.

I forgot the slash between 98 and sqrt(100)!. It was my typo. Sorry.
Still I get the same answer of z = -0.816 instead of -3.33. I am only concerned about getting the z value because how can I answer yes or no, without knowing it.
Also, it is asking for us to test using alpha = .05. Z-test is the only test that seems to apply here since we were are comparing a sample (California) from a population (National).
I also forgot to write the problem statement. I didn't notice it was omitted. See above in red.
 
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