9.73 Statistics on Large Sample Tests of Hypotheses?

  • Thread starter Little Bear
  • Start date
  • Tags
    Statistics
In summary, 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 does not 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.
  • #1
Little Bear
12
0

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.
 
Physics news on Phys.org
  • #2
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 [itex]98\sqrt{100}= 980[/itex], 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!)
 
Last edited by a moderator:
  • #3
Little Bear said:

Homework Statement


:yuck: (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))]
:yuck: 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.

:yuck: 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.
 
Last edited:

1. What are large sample tests of hypotheses in statistics?

Large sample tests of hypotheses are statistical methods used to make inferences about a population based on a sample of data. They involve using a sample size that is considered "large" in comparison to the population size, and making assumptions about the population based on the sample.

2. How are large sample tests of hypotheses different from small sample tests?

The main difference between large sample tests and small sample tests is the sample size used. Large sample tests use a sample size that is considered large relative to the population, while small sample tests use a sample size that is small in comparison. Large sample tests also make different assumptions about the population and can provide more accurate results.

3. What is a hypothesis in statistics?

In statistics, a hypothesis is a statement or assumption about a population that is tested using data from a sample. It is used to make predictions or inferences about the population based on the sample data.

4. How do you conduct a large sample test of a hypothesis?

To conduct a large sample test of a hypothesis, you first need to determine the null and alternative hypotheses. Then, you need to choose an appropriate statistical test based on the type of data and question being asked. Next, you collect a large sample of data and calculate the test statistic. Finally, you compare the test statistic to a critical value and make a decision about the null hypothesis.

5. What are some common large sample tests of hypotheses in statistics?

Some common large sample tests of hypotheses include the z-test, t-test, and chi-square test. These tests are used for different types of data and research questions, such as testing for differences between means, proportions, or variances in a population. Other large sample tests include ANOVA, regression analysis, and correlation analysis.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
2
Views
4K
  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
7K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
27
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
905
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
Replies
3
Views
12K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
Back
Top