Question

According to the National Health and Nutrition Survey (NHANES) sponsored by the U.S. government, a random...

According to the National Health and Nutrition Survey (NHANES) sponsored by the U.S. government, a random sample of 712 males between 20 and 29 years of age and a random sample of 1,001 males over the age of 75 were chosen and the weight of each of the males were recorded (in kg). Do the data provide evidence that the younger male population weighs more (on average) than the older male population? Use “Y” for ages 20-29 and “S” for ages 75+. If was found that x̅Y = 83.4, sY = 18.7, x̅S = 78.5, sS = 19.0.

Suppose the calculated confidence interval was (3.546, 6.254). What is the correct interpretation of this confidence interval?

  1. We are 95% confident that the true mean difference between the weight of males 20-29 and the weight of males 75+ lies between 3.546 and 6.254 kg.
  2. We are 95% confident that the true difference between the mean weight of males 20-29 and the mean weight of males 75+ lies between 3.546 and 6.254 kg.
  3. 95% of the time, the difference in the mean weight of males 20-29 and the mean weight of all males 75+ will lie between 3.546 and 6.254 kg.
  4. We are 95% confident that the true difference between the sample mean weight of males 20-29 and the same mean weight of males 75+ lies between 3.546 and 6.254.

A university administrator sampled the academic records of both male and female scholarship athletes at her university. After comparing the mean GPA of males with the mean GPA of females, she reported no significant difference between the means (p-value = 0.287). What is a correct interpretation of this p-value in the context of the problem?​​​​​​​

  1. The probability that the mean GPA of all male athletes equals the mean GPA of all female athletes is 0.287.
  2. The probability that the mean GPA of all male athletes differs from the mean GPA of all female athletes is 0.287.
  3. If the mean GPA of all male athletes is the same as the mean GPA for all female athletes, the probability of observing a difference between the sample mean GPA for male athletes and the sample mean GPA for female athletes as large or larger than that observed is 0.287.
  4. The probability that the mean GPA of the sample of male athletes differs from the mean GPA of the sample of female athletes is 0.287.

Using the confidence interval (3.546, 6.254), what would be the correct decision to make for the test of hypotheses with H0: μY = μS vs. Ha: μY ≠ μS at α = 0.05?​​​​​​​

  1. Because 0 is not included in the interval, we fail to reject the null hypothesis that the means are equal.
  2. Because 0 is included in the interval, we fail to reject the null hypothesis that the means are equal.
  3. Because 0 is not included in the interval, we reject the null hypothesis and conclude that the means are different.
  4. Because 0 is included in the interval, we reject the null hypothesis and conclude that the means are different.

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
1.) According to the National Health and Nutrition Survey (NHANES), the mean height of males ages...
1.) According to the National Health and Nutrition Survey (NHANES), the mean height of males ages 20 and over is 69.2 inches. I randomly selected 35 golfers from the PGA Tour money list. Their average height was 71.3 inches with a standard deviation of 2.2 inches. a.) At a ? = .05 significance level, can we conclude that on average Money List golfers are generally taller than the average male? b.) Give a 95% confidence interval for the mean height...
According to a recent survey by the National Health and Nutrition Examination Survey, the mean height...
According to a recent survey by the National Health and Nutrition Examination Survey, the mean height of adult men in the United States is 69.7 inches with a standard deviation of 3 inches. A sociologist believes that taller men may be more likely to be promoted to positions of leadership, so the mean height of male business executives would be greater than the mean heigth of of the entire male population. A SRS of of 100 male business executives has...
A park manager wants to determine if the amount of support for the national park is...
A park manager wants to determine if the amount of support for the national park is different among community members. More specifically, he/she wants to know if there is a difference between males and females. 1. What could be the potential research question for this problem? 2. What could be the hypothesis and alternative hypothesis for this problem? 3. Looking at the end-results of the T-Test that are here below, especially the MEAN of females and males, does it REJECT...
Exhibit 10-1 Salary information regarding two independent random samples of male and female employees of a...
Exhibit 10-1 Salary information regarding two independent random samples of male and female employees of a large company is shown below. Male Female Sample size 64 36 Sample mean salary (in $1000s) 44 41 Population variance 128 72 Refer to Exhibit 10-1. At 95% confidence, we have enough evidence to conclude that the  _____. a. We fail to reject the null hypothesis; we conclude that the average average salary of males is at least as much as females. b. We reject...
Suppose it is desired to compare the proportion of male and female students who voted in...
Suppose it is desired to compare the proportion of male and female students who voted in the last presidential election. We decide to randomly and independently sample 1000 male and 1000 female students and ask if they voted or not. A printout of the results is shown below. Hypothesis Test - Two Proportions Sample Size Successes Proportion Males 1000 475 0.47500 Females 1000 525 0.52500 Difference -0.05000 Null Hypothesis: p1 = p2 Alternative Hyp: p1 ≠ p2 SE (difference) 0.02236...
Salary information regarding male and female employees of a large company is shown below. Male Female...
Salary information regarding male and female employees of a large company is shown below. Male Female Sample Size: 64 36 Sample Mean Salary (in $1,000): 44 41 Population Variance:    128 72 1.) The standard error for the difference between the two means is 2.) The point estimate of the difference between the means of the two populations is 3.) At 95% confidence, the margin of error is 4.)  The 95% confidence interval for the difference between the means of the two...
The China Health and Nutrition Survey aims to examine the effects of the health, nutrition, and...
The China Health and Nutrition Survey aims to examine the effects of the health, nutrition, and family planning policies and programs implemented by national and local governments. It, for example, collects information on number of hours Chinese parents spend taking care of their children under age 6. The side-by-side box plots below show the distribution of this variable by educational attainment of the parent. Also provided below is the ANOVA output for comparing average hours across educational attainment categories. Df...
A newspaper reports that the average expenditure on Valentine's Day is $100.89. Do male and female...
A newspaper reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 40 male consumers was $136.67, and the average expenditure in a sample survey of 30 female consumers was $66.64. Based on past surveys, the standard deviation for male consumers is assumed to be $35, and the standard deviation for female consumers is assumed to be $20. (a) What is...
A newspaper reports that the average expenditure on Valentine's Day is $100.89. Do male and female...
A newspaper reports that the average expenditure on Valentine's Day is $100.89. Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 40 maleconsumers was $136.67, and the average expenditure in a sample survey of 30 female consumers was $66.64. Based on past surveys, the standard deviation for male consumers is assumed to be $45, and the standard deviation for female consumers is assumed to be $20. (a) What is the...
The average expenditure on Valentine's Day was expected to be $100.89 (USA Today, February 13, 2006)....
The average expenditure on Valentine's Day was expected to be $100.89 (USA Today, February 13, 2006). Do male and female consumers differ in the amounts they spend? The average expenditure in a sample survey of 50 male consumers was $135.13 and the average expenditure in a sample survey of 39 female consumers was $65.59. Based on past surveys, the standard deviation for male consumers is assumed to be $31 , and the standard deviation for female consumers is assumed to...