Correction on 9.73 Statistics on Large Sample Tests of Hypotheses?

So, using a z test and alpha = .05, we can solve the problem as follows:In summary, the study examines the college readiness of California high school students in comparison to students nationwide, as measured by their SAT scores. A sample of 100 California students from the class of 2003 were randomly selected and their verbal and math scores were recorded. The national average scores for the class of 2003 were 507 on the verbal portion and 519 on the math portion. The data provided is used to determine if the average verbal and math scores for all California students in the class of 2003 are different from the national average, and if there is a difference between the average math and verbal scores for all California students in
  • #1
Little Bear
12
0
1. Homework Statement
(I forgot to write out the problem statement in previous post)
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.

2. 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

3. The Attempt at a Solution
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 z = -3.33. Please show me how step-by-step to derive at the correct answer. Thank you very much.

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). The textbook is expecting students to know which tests (z-test, t-test, etc) to solve the problem. Also, the back of the book even says z = -3.33
 
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  • #2
Book's answer for part a is wrong on two accounts. You should use a t test; and the answer isn't yes. The reason for using a t test is because the standard deviation is not known, but estimated from the sample. In this case, the "z" statistic in fact follows a t distribution. However, with a large sample size (e.g. n > 30), the t distribution becomes close enough to the standard normal distribution that the z test can be used.
 
  • #3
for part a., but answer is yes.

I would like to clarify some issues with this problem and its solution. Firstly, the problem statement is not clear as it does not specify the type of test to be used (one-tailed or two-tailed) and the direction of the alternative hypothesis (greater than, less than, or not equal to). This can result in confusion and different interpretations of the problem.

Secondly, the data provided in the problem is not sufficient to perform a z-test as it does not include the population standard deviation. Without this information, we cannot accurately calculate the z-value.

Thirdly, the solution provided in the back of the book is incorrect. The z-value for part a should be -0.816, as both of us have calculated. This means that there is not enough evidence to conclude that the average verbal score for all California students is different from the national average.

In order to accurately answer the questions in this problem, we would need to perform a t-test instead of a z-test. This is because we do not have the population standard deviation and the sample size is relatively small (n=100). Additionally, we need to specify the type of test and direction of the alternative hypothesis in order to properly interpret the results.

To answer part a, we would need to set up the null and alternative hypotheses as follows:

Null hypothesis (H0): The average verbal score for all California students in the class of 2003 is equal to the national average of 507.
Alternative hypothesis (Ha): The average verbal score for all California students in the class of 2003 is different from the national average of 507.

We would then need to calculate the t-value using the formula t = (x̄ - μ) / (s/√n), where x̄ is the sample mean, μ is the population mean, s is the sample standard deviation, and n is the sample size. Plugging in the values from the problem, we get:

t = (499 - 507) / (98/√100) = -0.816

Since we are using a two-tailed test with a significance level of α = 0.05, we would need to compare this t-value to the critical t-value with 99 degrees of freedom (n-1). Using a t-table or a calculator, we can find that the critical t-value is ±1.984.

Since
 

1. What is the purpose of conducting a large sample test of hypotheses?

A large sample test of hypotheses is used to determine whether a sample statistic is significantly different from a hypothesized population parameter. This is important in drawing conclusions and making decisions based on data.

2. How is a large sample test of hypotheses different from a small sample test?

A large sample test is used when the sample size is large (usually greater than 30) and the population variance is known or can be estimated. In contrast, a small sample test is used when the sample size is small and the population variance is unknown.

3. What is the significance level in a large sample test of hypotheses?

The significance level, also known as alpha, is the probability of rejecting the null hypothesis when it is actually true. It is typically set at 0.05 or 0.01, but can vary depending on the study and its goals.

4. What is a p-value and how is it interpreted in a large sample test of hypotheses?

A p-value is the probability of obtaining a test statistic at least as extreme as the one observed, assuming the null hypothesis is true. In a large sample test, a p-value less than or equal to the significance level indicates that the null hypothesis can be rejected in favor of the alternative hypothesis.

5. What are the assumptions of a large sample test of hypotheses?

The main assumptions of a large sample test include a random and representative sample, independence of observations, and a normal distribution of the population. Additionally, the sample size should be sufficiently large (usually greater than 30) and the population variance should be known or estimated.

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