Reference no: EM131759776
Problem 1 - TRUE or FALSE
Explain your answer for full credit.
(a) In a normal distribution, 68% of the area under the curve is within one standard deviation of the mean.
(b) Given that a normal distribution has a mean of 50 and standard deviation of 5. The median is also 50.
(c) If the variance from a set of data is zero, then all of the data values must be identical.
(d) A 95% confidence interval is wider than and 99% confidence interval of the same parameter.
(e) It is easier to reject the null hypothesis if we have a smaller significance level.
Problem 2 -
The patient recovery time from a surgical procedure is normally distributed with a mean of 5.3 days and standard deviation of 2.1 days. HINT: See Illowsky, Chapter 6.
(a) What is the probability of spending more than 2 days in recovery after a surgical procedure?
(b) What is the 90th percentile for recovery time after a surgical procedure?
Problem 3 -
Human heights are known to be normally distributed. Men have a mean height of 70 inches and females a mean height of 64 inches. Both have a population standard deviation of 3 inches.
(a) Find the 1st Quartile (Q1) of the female height distribution
(b) Find the height of a female in the 90th percentile
(c) What is the probability that a randomly selected female is above 70 inches height?
Assume a random sample of 100 males is selected.
(d) What is the standard deviation of the sample mean (also known as the standard error of the mean)?
(e) What is the probability that the mean height of samples is less than 68 inches tall?
HINT: Since we know the population standard deviation, you will always use the Normal Distribution.
Problem 4 -
The average height of students has a normal distribution with a standard deviation of 2.5 inches. You want to estimate the mean height of students at your college to within 1-inch with 95% confidence. How many students should you measure? HINT: Be sure to consider the two-tail probability when evaluating the z-score for the confidence level. See Illowsky, Chapter 8, Page 427.
Problem 5 -
A random sample of 100 test scores has a sample mean of 85. Assume that the test scores have a population standard deviation of 10. Construct a 95% confidence interval estimate of the mean test scores. HINT: See the Illowsky textbook, Section 8.1.
Problem 6 -
A statistics instructor believes that 50% of his students are male. One of his students thinks there are more males than females taking statistics, so he decides to conduct a random survey of past courses. He found that 83 of 150 students were male. Answer the questions below. HINT: This is a test of a single population proportion. See Example 9.17 on page 489 of the Illowsky textbook.
(a) What is the null hypothesis (Ho) and what is the alternate hypothesis (Ha)?
(b) What is the standard error of the proportion (as Lane calls it) or the standard deviation of the single population proportion (as Illowsky calls it)?
(c) What is the test statistic, or in this case, the z-score?
(d) What is the P-value for this test? HINT: Assume a normal distribution and use the single population proportion equations on page-504 of Illowsky (Mean = p, Standard Deviation = sqrt ( p * q / n).
(e) Is there sufficient evidence to reject the null hypothesis (Ho) at a 99% confidence level (α = 0.01)?
Problem 7 -
Taking an SAT preparatory course is believed to improve test scores. To investigate the effects of taking a preparatory course, scores were recorded for 5 students before and after they took the preparatory course. Does the data below suggest that a preparatory course increases test scores? Assume we want a 90% confidence level (?=0.10) to test the claim. HINT: Assume the null hypothesis is that test scores stay the same, before and after taking the course. Also, assume the "expected" change in test scores is zero when evaluating the test statistic. To find the test statistics, you should create a new column for the "change in test score" and evaluate the mean and standard deviation of those results.
|
Student
|
Before
|
After
|
|
1
|
1200
|
1260
|
|
2
|
1100
|
1060
|
|
3
|
1050
|
950
|
|
4
|
1000
|
1000
|
|
5
|
920
|
1050
|
(a) Identify the alternate hypothesis
(b) What is the random variable (X) that we want to evaluate?
(c) Find the test statistic (show all work)
(d) Find the P-value (show all work)
(e) Is there sufficient evidence to support the claim that taking a preparatory course increases an SAT test score? Justify your conclusion.
Problem 8 -
Researchers interviewed young adults in Canada and the U.S to compare their ages when they enter the workforce. The mean age of the 100 Canadians was 18 with a standard deviation of 6. The mean age of the 130 people interviewed in the U.S. was 20 with a standard deviation of 8. Is the mean age of entering the workforce in Canada lower than the mean age in the U.S.?
(a) Identify the null and alternate hypothesis.
(b) Find the test statistic (show all work). HINT: See Section 10.1 of Illowsky. You are testing two population means with unknown population standard deviations (we approximate the population mean with the sample mean). See the equation for the test statistic (t-score) on page-556 of Illowsky.
(c) Find the P-value (show all work) and evaluate whether the mean age of entering the workforce in Canada is lower than the mean age in the U.S.? Test at a 1% significance level. HINT: You should assume the underlying population is normally distributed, even though we are working with samples. This means you DO NOT have to find the "degrees of freedom" and you will use your test statistic as a "z-score" in a Normal Distribution. So, in EXCEL, you will use the NORMSDIST function.