TIMSS 2003 (Trends in International Mathematics and Science Study) focused on th
ID: 3134101 • Letter: T
Question
TIMSS 2003 (Trends in International Mathematics and Science Study) focused on the mathematics and science achievement of eighth-graders throughout the world. A total of 45 countries (including the United States) participated in the study. The mean math exam score for U.S. students was 512 with a standard deviation of 91. Assume the scores are normally distributed and a sample of 140 students is taken. (Give your answers correct to four decimal places.)
(a) Find the probability that the mean TIMSS score for a randomly selected group of eighth-graders would be between 495 and 519.
(b) Find the probability that the mean TIMSS score for a randomly selected group of eighth-graders would be less than 511.
(c) Do you think the assumption of normality is reasonable? Explain.
Yes, since the sample size is large.
Yes, since the sample size is small.
No, since the sample size is large.
No, since the sample size is small.
You may need to use the appropriate table in Appendix B to answer this question
Question 2
A random sample of size 44 is to be selected from a population that has a mean = 45 and a standard deviation of 10.
(a) This sample of 44 has a mean value of x, which belongs to a sampling distribution. Find the shape of this sampling distribution.
1.skewed right
2.approximately normal
3.skewed left
4.chi-square
(b) Find the mean of this sampling distribution. (Give your answer correct to nearest whole number.)
(c) Find the standard error of this sampling distribution. (Give your answer correct to two decimal places.)
(d) What is the probability that this sample mean will be between 40 and 55? (Give your answer correct to four decimal places.)
(e) What is the probability that the sample mean will have a value greater than 54? (Give your answer correct to four decimal places.)
(f) What is the probability that the sample mean will be within 2 units of the mean? (Give your answer correct to four decimal places.)
You may need to use the appropriate table in Appendix B to answer this question.
Question 3
A sample of 51 night-school students' ages is obtained in order to estimate the mean age of night-school students. x = 24.6 years. The population variance is 22.
(a) Give a point estimate for . (Give your answer correct to one decimal place.)
(b) Find the 95% confidence interval for . (Give your answer correct to two decimal places.)
(c) Find the 99% confidence interval for . (Give your answer correct to two decimal places.)
You may need to use the appropriate table in Appendix B to answer this question.
Question 4
The sampled population is normally distributed, with the given information. (Give your answers correct to two decimal places.)
n = 12, x = 29.2, and = 6.6
(a) Find the 0.9 confidence interval for .
(b) Are the assumptions satisfied? Explain.
Yes, the sampled population is normally distributed.
No, the sample distribution is not normally distributed.
not enough information
You may need to use the Z table to answer this question.
Question 5
Consider the following. (Give your answers correct to two decimal places.)
(a) Determine the value of the confidence coefficient z(/2) for 1 - = 0.83.
(b) Determine the value of the confidence coefficient z(/2) for 1 - = 0.98.
Question 6
In a study of 28 criminals convicted of antitrust offenses, the average age was 58 years, with a standard deviation of 7.1 years. Construct a 98% confidence interval for the true mean age of these type of offenders in the general population. (Give your answers correct to one decimal place.)
what to what years?
Question 7
Identify each numerical value by "name" (e.g., mean, variance) and by symbol (e.g.,x).
(a) The mean height of 24 junior high school girls is 4'11".
24 = sample size = n; 4'11" = sample mean = x
24 = sample mean = x ; 4'11" = sample size = n
24 = sample variance = s2; 4'11" = sample size = n
24 = population variance = 2; 4'11" = population mean =
(b) The standard deviation for IQ scores is 16.
16 = sample mean = x
16 = population standard deviation =
16 = sample variance = s2
16 = sample size = n
(c) The variance among the test scores on last week's exam was 190.
190 = sample variance = s2
190 = population mean =
190 = sample size = n
190 = population standard deviation =
(d) The mean height of all cadets who have ever entered West Point is 69 inches.
69 = sample size = n
69 = population mean =
69 = population standard deviation =
69 = sample mean = x
You may need to use the Unit Normal or Z table to answer this question.
Lower Limit Upper LimitExplanation / Answer
1.
TIMSS 2003 (Trends in International Mathematics and Science Study) focused on the mathematics and science achievement of eighth-graders throughout the world. A total of 45 countries (including the United States) participated in the study. The mean math exam score for U.S. students was 512 with a standard deviation of 91. Assume the scores are normally distributed and a sample of 140 students is taken. (Give your answers correct to four decimal places.)
A)
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 495
x2 = upper bound = 519
u = mean = 512
n = sample size = 140
s = standard deviation = 91
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -2.210403435
z2 = upper z score = (x2 - u) * sqrt(n) / s = 0.91016612
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.013538588
P(z < z2) = 0.818632546
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.805093958 [ANSWER]
**********************
b)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 511
u = mean = 512
n = sample size = 140
s = standard deviation = 91
Thus,
z = (x - u) * sqrt(n) / s = -0.130023731
Thus, using a table/technology, the left tailed area of this is
P(z < -0.130023731 ) = 0.448273826 [ANSWER]
***********************
c)
We usually need a sample size n > 30 to assume normality. As n = 140 here, n > 30. Hence,
OPTION A: Yes, since the sample size is large. [ANSWER]
*******************************************
Hi! Please submit the next part as a separate question. That way we can continue helping you! Please indicate which parts are not yet solved when you submit. Thanks!
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.