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I'm still working on this problem. I won't finish it tonight. My brain is MUSH....My ENTIRE weekends are spent trying to self teach myself this stuff!
#2
a. 0.488 <p< 0.505
b. NO, because the proportion could easily....
#4
E - Neither normal nor t distribution apples
I still haven't been able to figure out number 3 and 5. ---and #9 on a different post...
ANY help is greatly appreciated...
Thanks
For this one I got 0.086 <p< 0.132
Wonder what I did wrong?
IM A DING BAT....I DIDNT GET IT WRONG...I GOT THE CORRECT ANSWER....
#6
I got.... 0.119 < u < 0.181
and... D-NO IT IS POSSIBLE THAT THE REQUIREMENT IS BEING MET.....
#7
833
and D - there are no obstacles to getting a good....
#8
I got.....B-No the presence of a linear correlation between 2 variables ........
Anyone care to help me?
Use the given data set to complete parts (a) through (c) below. (Use α =0.05.)
x 10 8 13 9 11 14 6 4 12 7 5
y 9.15 8.15 8.73 8.77 9.26 8.11 6.13 3.11 9.13 7.26 4.73
b. Find the linear correlation coefficient, r, then determine whether there is sufficient evidence to support the claim of a linear correlation between the two variables.
The linear correlation coefficient is r = ______. (Round to three decimal places as needed.)
Determine whether there is sufficient evidence to support the claim of a linear correlation between the two variables. Choose the correct answer below.
A. There is insufficient evidence to support the claim of a linear correlation between the two variables.
B. There is sufficient evidence to support the claim of a linear correlation between the two variables.
C. There is insufficient evidence to support the claim of a nonlinear correlation between the two variables.
D. There is sufficient evidence to support the claim of a nonlinear correlation between the two variables.
c. Identify the feature of the data that would be missed if part (b) was completed without constructing the scatterplot. Choose the correct answer below.
A. The scatterplot reveals a distinct pattern that is not a straight-line pattern.
B. The scatterplot reveals a distinct pattern that is a straight-line pattern with negative slope.
C. The scatterplot does not reveal a distinct pattern.
D. The scatterplot reveals a distinct pattern that is a straight-line pattern with positive slope.
10.
For a sample of eight bears, researchers measured the distances around the bears' chests and weighed the bears. Minitab was used to find that the value of the linear correlation coefficient is r=0.889. Using α=0.05, determine if there is a linear correlation between chest size and weight. What proportion of the variation in weight can be explained by the linear relationship between weight and chest size?
a. Is there a linear correlation between chest size and weight?
A. Yes, because the absolute value of the test statistic exceeds the critical value of 0.707.
B. No, because the absolute value of the test statistic exceeds the critical value of 0.707.
C. Yes, because the test statistic falls between the critical values of −0.707 and 0.707.
D. The answer cannot be determined from the given information.
b. What proportion of the variation in weight can be explained by the linear relationship between weight and chest size?
___________(Round to three decimal places as needed.)
6.
In a sample of seven cars, each car was tested for nitrogen-oxide emissions (in grams per mile) and the following results were obtained: 0.16, 0.12, 0.17, 0.11, 0.15, 0.16, 0.18. Assuming that this sample is representative of the cars in use, construct a 98% confidence interval estimate of the mean amount of nitrogen-oxide emissions for all cars. If the EPA requires that nitrogen-oxide emissions be less than 0.165 g/mi, can we safely conclude that this requirement is being met?
What is the confidence interval estimate of the mean amount of nitrogen-oxide emissions for all cars?
_______g/mi<μ<______ g/mi (Round to three decimal places as needed.)
Can we safely conclude that the requirement that nitrogen-oxide emissions be less than 0.165 g divided by mig/mi is being met?
A. No, because the confidence interval does not contain 0.165 g/mi.
B. Yes, we can definitely conclude that the requirement is met for all cars.
C. Yes, because the confidence interval contains 0.165 g/mi.
D. No, it is possible that the requirement is being met, but it is also very possible that the mean is not less than 0.165 g/mi.
7.
In order to estimate the mean amount of time computer users spend on the internet each month, how many computer users must be surveyed in order to be 90% confident that your sample mean is within 13 minutes of the population mean? Assume that the standard deviation of the population of monthly time spent on the internet is 228 min. What is a major obstacle to getting a good estimate of the population mean? Use technology to find the estimated minimum required sample size.
The minimum sample size required is ________ computer users. (Round up to the nearest whole number.)
What is a major obstacle to getting a good estimate of the population mean?
A. There may not be 833 computer users to survey.
B. The data does not provide information on what the computer users did while on the internet.
C. It is difficult to precisely measure the amount of time spent on the internet, invalidating some data values.
D. There are no obstacles to getting a good esitmate of the population mean.
8.
If we find that there is a linear correlation between the concentration of carbon dioxide in our atmosphere and the global temperature, does that indicate that changes in the concentration of carbon dioxide cause changes in the global temperature?
Choose the correct answer below.
A. Yes. The presence of a linear correlation between two variables implies that one of the variables is the cause of the other variable.
B. No. The presence of a linear correlation between two variables does not imply that one of the variables is the cause of the other variable.
Many thanks to anyone that helps me. This is not something I do to copy....I do my own work as I have to turn in all the steps worked out and I like to compare my answers because I know very very little about this stuff. I am only taking this class because I must to proceed in my graduate class. I am not a math student, I am a nurse. It is a required class and it has been 18 years since I've had any math and it is online so I am having to self-teach all of it to myself. Thanks for your help.
1.
Express the confidence interval (0.086, 0.132) in the form of
p−E <p <p+E.
________<p<________(Type integers or decimals).
2.
In the week before and the week after a holiday, there were 10,000 total deaths, and 4963 of them occurred in the week before the holiday.
a. Construct a 90% confidence interval estimate of the proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday.
b. Based on the result, does there appear to be any indication that people can temporarily postpone their death to survive the holiday?
a. _______<p<__________(Round to three decimal places as needed.)
b. Based on the result, does there appear to be any indication that people can temporarily postpone their death to survive the holiday? (no or yes)
No, because the proportion could easily equal 0.5. The interval is not less than 0.5 the week before the holiday.
Yes, because the proportion could not easily equal 0.5. The interval isis substantially less than 0.5 the week before the holiday.
3.
An online site presented this question, 'Would the recent norovirus outbreak deter you from taking a cruise?' Among the 34,932 people who responded, 68% answered 'yes'. Use the sample data to construct a 95% confidence interval estimate for the proportion of the population of all people who would respond 'yes' to that question. Does the confidence interval provide a good estimate of the population proportion?
_______< p <_______(Round to three decimal places as needed.)
Does the confidence interval provide a good estimate of the population proportion?
A.No, the sample is a voluntary sample and might not be representative of the population.
B.No, the responses are not independent.
C.Yes, the sample is large enough to provide a good estimate of the population proportion.
D.Yes, all the assumptions for a confidence interval are satisfied.
4.
Do one of the following, as appropriate. (a) Find the critical value z α/2, (b) find the critical value t α/2, (c) state that neither the normal nor the t distribution applies.
Confidence level 95%; n=23; σ is known; population appears to be very skewed.
A. z α/2 =1.645
B. z α/2 =1.96
C. t α/2 =1.717
D. t α/2 =2.074
E. Neither normal nor t distribution applies.
5.
A data set includes 103 body temperatures of healthy adult humans for which x = 98.1degrees°F and s=0.56°F. Complete parts (a) and (b) below.
a. What is the best point estimate of the mean body temperature of all healthy humans?
The best point estimate is ________degrees°F. (Type an integer or a decimal.)
b. Using the sample statistics, construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. Do the confidence interval limits contain 98.6°F? What does the sample suggest about the use of 98.6 °F as the mean body temperature?
What is the confidence interval estimate of the population mean μ?
_______°F < μ < ________°F (Round to three decimal places as needed.)
Do the confidence interval limits contain 98.6 °F? (no or yes)
________No
________Yes
What does this suggest about the use of 98.6 °F as the mean body temperature?
A.This suggests that the mean body temperature could be higher than 98.6 °F.
B.This suggests that the mean body temperature could be lower than 98.6 °F.
C.This suggests that the mean body temperature could be very possibly be 98.6 °F.
bob..can you help me set up 4c
Thank you for your help....SO THANKFUL MY CAREER IS NURSING AND NOT THIS MATH...I do not understand all this stuff!!!!!
Can you tell me how to set the problems for #4 and #8 up??
Thanks!
Bobbym,
I need this one more than 7b. I don't even know how to set this one up . Can you help?
8. A certain brand of candies have a mean weight of 0.8602 g and a standard deviation of 0.0511 A sample of these candies came from a package containing 464 candies, and the package label stated that the net weight is 396.4 g. (If every package has 464 candies, the mean weight of the candies must exceed 396.4/ 464 (Fraction)=0.8543 g for the net contents to weigh at least 396.4 g
a. If 1 candy is randomly selected, find the probability that it weighs more than 0.8543 g.
The probability is _________. (Round to four decimal places as needed.)
b. If 464 candies are randomly selected, find the probability that their mean weight is at least 0.8543 g.
The probability that a sample of 464 candies will have a mean of 0.8543 g or greater is _________.(Round to four decimal places as needed.)
c. Given these results, does it seem that the candy company is providing consumers with the amount claimed on the label?
▼ Yes or No_________because the probability of getting a sample mean of 0.8543 g or greater when 464 candies are selected
▼ is or is not_________ exceptionally small.
Thanks!
Like so close that you think my answer is correct or you think I should rework number 7 and 10? What did you get for 7b?
what about this one?
8. A certain brand of candies have a mean weight of 0.8602 g and a standard deviation of 0.0511 A sample of these candies came from a package containing 464 candies, and the package label stated that the net weight is 396.4 g. (If every package has 464 candies, the mean weight of the candies must exceed 396.4/ 464 (Fraction)=0.8543 g for the net contents to weigh at least 396.4 g
a. If 1 candy is randomly selected, find the probability that it weighs more than 0.8543 g.
The probability is _________. (Round to four decimal places as needed.)
b. If 464 candies are randomly selected, find the probability that their mean weight is at least 0.8543 g.
The probability that a sample of 464 candies will have a mean of 0.8543 g or greater is _________.(Round to four decimal places as needed.)
c. Given these results, does it seem that the candy company is providing consumers with the amount claimed on the label?
▼ Yes or No_________because the probability of getting a sample mean of 0.8543 g or greater when 464 candies are selected
▼ is or is not_________ exceptionally small.
Thanks for your help
What did you get for 4---- Letter c?
#7....this is what i got
a. 0.6255
b. 0.9756
#10...this is what i got
0.7157
Did you get the same thing for #7 and #10? and also let me know when you get the others above.
Thanks!
Tonnie
I DO THE PROBLEMS OVER AND OVER AND THEN ITS LIKE I CANNOT FIGURE OUT HOW TO SET THEM UP AND THEN I QUESTION MY ANSWERS AND OF COARSE IF ONE LITTLE THING IS OFF THEN IT IS.....WRONG. CAN YOU HELP ME WITH THESE SO I CAN COMPARE WHAT I GET? THANKS!!!
4. A survey found that women's heights are normally distributed with mean 62.4in. and standard deviation 2.5 in. The survey also found that men's heights are normally distributed with a mean 69.1 in. and standard deviation 2.9. Complete parts a through c below.
a. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in. and a maximum of 6 ft 3 in. Find the percentage of women meeting the height requirement.
The percentage of women who meet the height requirement is ________%.
(Round to two decimal places as needed.)
b. Find the percentage of men meeting the height requirement.
The percentage of men who meet the height requirement is _________%.
(Round to two decimal places as needed.)
c. If the height requirements are changed to exclude only the tallest 5% of men and the shortest 5% of women, what are the new height requirements?
The new height requirements are at least ________ in. and at most __________ in.
(Round to one decimal place as needed.)
6. The population of current statistics students has ages with mean μ and standard deviation σ. Samples of statistics students are randomly selected so that there are exactly 53 students in each sample. For each sample, the mean age is computed. What does the central limit theorem tell us about the distribution of those mean ages?
Choose the correct answer below.
A. Because n>30, the sampling distribution of the mean ages can be approximated by a normal distribution with mean μ and standard deviation σ divided by square root of 53.
B. Because n>30, the sampling distribution of the mean ages is precisely a normal distribution with mean μ and standard deviation σ/(divided by) square root of 53.
C. Because n>30, the sampling distribution of the mean ages can be approximated by a normal distribution with mean μ and standard deviation σ.
D.Because n>30, the central limit theorem does not apply in this situation.
7. Assume that women's heights are normally distributed with a mean given by μ=62.4 in, and a standard deviation given by σ=1.9 in.
(a) If 1 woman is randomly selected, find the probability that her height is less than 63 in.
(b) If 39 women are randomly selected, find the probability that they have a mean height less than 63 in.
(a) The probability is approximately _________.
(Round to four decimal places as needed.)
(b) The probability is approximately _________.
(Round to four decimal places as needed.)
8. A certain brand of candies have a mean weight of 0.8602 g and a standard deviation of 0.0511 A sample of these candies came from a package containing 464 candies, and the package label stated that the net weight is 396.4 g. (If every package has 464 candies, the mean weight of the candies must exceed 396.4/ 464 (Fraction)=0.8543 g for the net contents to weigh at least 396.4 g
a. If 1 candy is randomly selected, find the probability that it weighs more than 0.8543 g.
The probability is _________. (Round to four decimal places as needed.)
b. If 464 candies are randomly selected, find the probability that their mean weight is at least 0.8543 g.
The probability that a sample of 464 candies will have a mean of 0.8543 g or greater is _________.(Round to four decimal places as needed.)
c. Given these results, does it seem that the candy company is providing consumers with the amount claimed on the label?
▼ Yes or No_________because the probability of getting a sample mean of 0.8543 g or greater when 464 candies are selected
▼ is or is not_________ exceptionally small.
10. Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that 144 eligible voters aged 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted.
Probability that fewer than 35 voted
The probability that fewer than 35 of 144 eligible voters voted is __________.(Round to four decimal places as needed.)
Thanks!!!! Ill keep trying!
I got the question wrong. said the answer was 0.000 so now i am REALLY CONFUSED!
I don't think I understand. I am getting something very different. This is what I am doing. I am seeing that this is a cumulative binomial probability
distribution. Because it states that "no more than" 1. we have to solve for P(x=0) and P(x=1) and then add those together.
P= 0.75
N=9
X=1 p(x=0) = 3.814697266
Q=0.25 p(x=1) = 1.0299682 added together being 4.845 being the probability of no more than 1.
setting it up like
p(x)= n!/(n-x)!x! multiplied by p to the x power and multiplied by q to the n-x power.
do this formula twice once with x equally 0 and once with x=1 and them the sum would be the probability of "no more than" 1 of the offspring peas has a green pod.
This is all new to me so I know I could be WAY OFF but this is what I thought it was. Am I wrong?
Nine peas are generated from parents having the green/yellow pair of genes, so there is a 0.75 probability that an individual pea will have a green pod. Find the probability that among the 9 offspring peas, no more than 1 has a green pod. Is it unusual to get no more than 1 pea with a green pod when 9 offspring peas are generated? Why or why not?
The probability that no more than 1 of the 9 offspring peas has a green pod is ___________.
(Round to three decimal places as needed.)
Is it unusual to randomly select 9 peas and find that no more than 1 of them have a green pod? Note that a small probability is one that is less than 0.05.
A. Yes, because the probability of this occurring is not small.
B. Yes, because the probability of this occurring is very small.
C. No, because the probability of this occurring is not small.
D. No, because the probability of this occurring is very small.
Can you help me with this? I cannot even figure out how to set it up let alone get the answer. PLEASE show me how to set it up and the answer. This problem is due tuesday. 1/26/16
Thank you in advance.