Question

In: Statistics and Probability

Suppose a simple random sample of athletes in the NBA heights is taken. There were 28...

Suppose a simple random sample of athletes in the NBA heights is taken. There were 28 athletes in the sample with a mean height of 78.4 inches and standard deviation of 2 inches. It has been confirmed through statistical analysis that NBA player heights follows a normal distribution.

a. what parameter are we estimating?

b. Explain the requirements as they relate to the problem

c.What is the point estimate of the parameter?

d.Input the margin of error for a 95% confidence interval for the true average height of NBA players.
Write the equation you used and the numbers .

e. Create and input your interval using your answer in part d.

f.  Interpret your interval. (explanation)

g.Use your interval to respond to the statement that the true average height of NBA players is less than than 78.4. Thoroughly explain why you are responding the way you are.

h. What does the central limit theorem say about the sample distribution of the sample average of basketball player heights for this problem?

please show all work and round to the fourth on all ansers

Solutions

Expert Solution

hii... although I am trying to provide the detailed answer but if you have any doubt please ask by comment. please like the answer. thanks..


Related Solutions

A random sample of 18 women is taken and their heights were recorded. The heights (in...
A random sample of 18 women is taken and their heights were recorded. The heights (in inches) are: 60, 62, 63, 63, 63, 66, 66, 66, 66, 67, 67, 68, 68, 68, 69, 70, 71, 71 Assume that women’s height are normally distributed. Let μ be the mean height of all women and let p be the proportion of all women that are taller than 54 inches. a) Test the hypothesis H0: μ = 54 against H1: μ > 54....
The heights were recorded for a Simple Random Sample of 270 junior.  The mean of this sample...
The heights were recorded for a Simple Random Sample of 270 junior.  The mean of this sample was 66.5 inches.  The heights are known to be Normally Distributed with a population standard deviation of 5.1 inches. (You do not need a data set for this). Test the claim that the mean height of Juniors has increased from 65.7 at a 0.01 significance level. (Use MINITAB to do the hypothesis test and copy and paste the output of the hypothesis test here (0.5pts)....
A simple random sample of 28 recent birth records at the local hospital was taken. In...
A simple random sample of 28 recent birth records at the local hospital was taken. In the sample, the average birth weight was 119.6 ounces and the sample standard deviation was 6.5 ounces. Assume that in the population of all babies born in this hospital, the birth weights follow a Normal distribution, with mean µ. Construct a 98% confidence interval for µ, the average birthweight of newborns. (Show work please)
A simple random sample was taken of 1000 shoppers Respondents were classified by gender (male or...
A simple random sample was taken of 1000 shoppers Respondents were classified by gender (male or female) and by meat department preference (beef, chicken, fish). Results are shown in the contingency table:         Gender: Beef Chicken Fish Row Total Male 200 150 50 400 Female 250 300 50 600 Column Total 450 450 100 1000 Is there a difference in meat preference between males and females? The solution to this problem takes four steps: (1) state the hypotheses, (2)...
A simple random sample was taken of 1000 shoppers Respondents were classified by gender (male or...
A simple random sample was taken of 1000 shoppers Respondents were classified by gender (male or female) and by meat department preference (beef, chicken, fish). Results are shown in the contingency table: Gender: Beef Chicken Fish Row Total Male 200 150 50 400 Female 250 300 50 600 Column Total 450 450 100 1000 Is there a difference in meat preference between males and females? The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an...
Pulse rates for Men A simple random sample of 40 pulse rates of women were taken....
Pulse rates for Men A simple random sample of 40 pulse rates of women were taken. The mean of women’s pulse rates have a mean of 67.3 beats per minute and a standard deviation of 10.3 beats per minute. Construct a 99% confidence interval estimate of the standard deviation of the pulse rates of men. Pulse rates are normally distributed.
1. A simple random sample of 25 items resulted in a sample mean of 28 and...
1. A simple random sample of 25 items resulted in a sample mean of 28 and a standard deviation of 7.5. Construct a 95% confidence interval for the population mean. Select one: A. 24.904 to 31.096 B. 25.905 to 32.096 C. 26.324 to 29.432 D. None of the above answers is correct 2. A simple random sample of 25 items resulted in a sample mean of 28 and a standard deviation of 7.5. Construct a 99% confidence interval for the...
The heights of a simple random sample of soccer players in a particular league are given...
The heights of a simple random sample of soccer players in a particular league are given below. Can you conclude at the 5% level of significance, that the average height of soccer players in the league sampled is over 182 cm? Assume that the heights of soccer players is normal. Show all of your work, include all necessary steps, and be complete in your answer and explanation. 193 190 185.3 193 172.7 180.3 186 188
The heights of a simple random sample of 400 male high school sophomores in a Midwestern...
The heights of a simple random sample of 400 male high school sophomores in a Midwestern state are measured. The sample mean is = 66.2 inches. Suppose that the heights of male high school sophomores follow a Normal distribution with a standard deviation of σ = 4.1 inches. What is a 95% confidence interval for µ? A. (59.46, 72.94) B. (58.16, 74.24) C. (65.80, 66.60) D. (65.86, 66.54)
A simple random sample of 400 persons is taken to estimate the percentage of Republicans in...
A simple random sample of 400 persons is taken to estimate the percentage of Republicans in a large population. It turns out that 210 of the people in the sample are Republicans. True or False and explain. a. The sample percentage is 52.5%; the SE for the sample percentage is 2.5%. b. 52.5% ± 2.5% is a 75%-confidence interval for the population percentage. c. 52.5% ± 5% is a 95%-confidence interval for the sample percentage. d. 52.5% ± 5% is...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT