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4Homework Answers
As HOMEWORKLIB RULES guideline expert have to answer first question only since I had done 2 for you please upvote
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means so much...
thank you for help! it means a lot to me ! and can u please underline...
thank you for help! it means a lot to me ! and can u please
underline or circle the answers. thank you
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(1 point) A random sample of 100 observations from a population with standard deviation 13.83 yielded a sample mean of 92.3. Given that the null hypothesis is u = 90 and the alternative hypothesis is u > 90 using a = .05, find the following: (a) Test statistic: (b) P-value: DOO The conclusion for this test...
thank you for all the help! i appreciate everythjng! 1 2 3 (1 point) A random...
thank you for all the help! i appreciate everythjng!
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(1 point) A random sample of 100 observations from a population with standard deviation 8.31 yielded a sample mean of 91.5. Part 1: Part 2: Part 3: Given that the null hypothesis is j = 90 and the alternative hypothesis is #90 using a = .05, find the following: (a) Test statistic = (b) P-value: Note: Round off the test statistic to 2 decimal places and the P-value...
(1 point) A random sample of 100 observations from a population with standard deviation 5.24 yielded...
(1 point) A random sample of 100 observations from a population with standard deviation 5.24 yielded a sample mean of 91. Part 1: Given that the null hypothesis is j = 90 and the alternative hypothesis is ju > 90 using : .05, find the following: (a) Test statistic = A = (b) P-value: Part 2
(1 point) A random sample of 100 observations from a population with standard deviation 25.72 yielded...
(1 point) A random sample of 100 observations from a population with standard deviation 25.72 yielded a sample mean of 94.3. Part 1: Given that the null hypothesis is u = 90 and the alternative hypothesis is j > 90 using a = .05, find the following: (a) Test statistic = (b) P-value: Note: Round off the test statistic to 2 decimal places and the P-value to 4 decimal places.
please help. thank you so much !! 1 2 0.0702 0.0702 incorrect incorrect At least one...
please help. thank you so much !!
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2
0.0702 0.0702 incorrect incorrect At least one of the answers above is NOT correct. 1 of the questions remains unanswered. (1 point) A random sample of 100 observations from a population with standard deviation 8.31 yielded a sample mean of 91.5. Part 1: Part 2: - Part 3 Given that the null hypothesis is p = 90 and the alternative hypothesis is x 90 using a = .05, find the following:...
Homework4: Problem 2 Previous Problem Problem List Next Problem (6 points) A random sample of 100...
Homework4: Problem 2 Previous Problem Problem List Next Problem (6 points) A random sample of 100 observations from a population with standard deviation 23.75 yielded a sample mean of 94.5. 1. Given that the null hypothesis is H = 90 and the alternative hypothesis is u > 90 using a = .05, find the following: (a) Test statistic = (b) P-value: (c) The conclusion for this test is: A. There is insufficient evidence to reject the null hypothesis B. Reject...
Ch6 Sec2: Problem 3 PreviousProblem List Next (1 point) A random sample of 100 observations from...
Ch6 Sec2: Problem 3 PreviousProblem List Next (1 point) A random sample of 100 observations from a population with standard deviation 9.13 yielded a sample mean of 91.6 1. Given that the null hypothesis is--90 and the alternative hypothesis is μ > 90 using a) Test statistic- (b) P- value: (c) The conclusion for this test is: 05, find the following: A. Reject the null hypothesis B. There is insufficient evidence to reject the null hypothesis C. None of the...
im so grateful for all your help! this has helped me in so many ways especially...
im so grateful for all your help! this has helped me in so
many ways especially with all the things going on right now. thank
you so much !!
1
2
3
4
(1 point) 35 people are randomly selected and the accuracy of their wristwatches is checked, with positive errors representing watches that are ahead of the correct time and negative errors representing watches that are behind the correct time. The 35 values have a mean of 94sec and...
In a sample of credit card holders the mean monthly value of credit card purchases was...
In a sample of credit card holders the mean monthly value of credit card purchases was $ 400 and the sample variance was 67 ($ squared). Assume that the population distribution is normal. Answer the following, rounding your answers to two decimal places where appropriate. (a) Suppose the sample results were obtained from a random sample of 12 credit card holders. Find a 95% confidence interval for the mean monthly value of credit card purchases of all card holders. ( , )...
THANK YOU IN ADVANCE FOR HELPING ME! i appreciate this soo much! this has helped me...
THANK YOU IN ADVANCE FOR HELPING ME! i appreciate this soo
much! this has helped me a lot especially throught times like this.
im so grateful for
your help
1
2
3
4
(1 point) The time needed for college students to complete a certain paper-and-pencil maze follows a normal distribution with a mean of 30 seconds and a standard deviation of 2.3 seconds. You wish to see if the mean time is changed by vigorous exercise, so you have...
thank you for help! it means a lot to me ! and can u please
underline or circle the answers. thank you
1
2
(1 point) A random sample of 100 observations from a population with standard deviation 13.83 yielded a sample mean of 92.3. Given that the null hypothesis is u = 90 and the alternative hypothesis is u > 90 using a = .05, find the following: (a) Test statistic: (b) P-value: DOO The conclusion for this test...
thank you for all the help! i appreciate everythjng!
1
2
3
(1 point) A random sample of 100 observations from a population with standard deviation 8.31 yielded a sample mean of 91.5. Part 1: Part 2: Part 3: Given that the null hypothesis is j = 90 and the alternative hypothesis is #90 using a = .05, find the following: (a) Test statistic = (b) P-value: Note: Round off the test statistic to 2 decimal places and the P-value...
(1 point) A random sample of 100 observations from a population with standard deviation 5.24 yielded a sample mean of 91. Part 1: Given that the null hypothesis is j = 90 and the alternative hypothesis is ju > 90 using : .05, find the following: (a) Test statistic = A = (b) P-value: Part 2
(1 point) A random sample of 100 observations from a population with standard deviation 25.72 yielded a sample mean of 94.3. Part 1: Given that the null hypothesis is u = 90 and the alternative hypothesis is j > 90 using a = .05, find the following: (a) Test statistic = (b) P-value: Note: Round off the test statistic to 2 decimal places and the P-value to 4 decimal places.
please help. thank you so much !!
1
2
0.0702 0.0702 incorrect incorrect At least one of the answers above is NOT correct. 1 of the questions remains unanswered. (1 point) A random sample of 100 observations from a population with standard deviation 8.31 yielded a sample mean of 91.5. Part 1: Part 2: - Part 3 Given that the null hypothesis is p = 90 and the alternative hypothesis is x 90 using a = .05, find the following:...
Homework4: Problem 2 Previous Problem Problem List Next Problem (6 points) A random sample of 100 observations from a population with standard deviation 23.75 yielded a sample mean of 94.5. 1. Given that the null hypothesis is H = 90 and the alternative hypothesis is u > 90 using a = .05, find the following: (a) Test statistic = (b) P-value: (c) The conclusion for this test is: A. There is insufficient evidence to reject the null hypothesis B. Reject...
Ch6 Sec2: Problem 3 PreviousProblem List Next (1 point) A random sample of 100 observations from a population with standard deviation 9.13 yielded a sample mean of 91.6 1. Given that the null hypothesis is--90 and the alternative hypothesis is μ > 90 using a) Test statistic- (b) P- value: (c) The conclusion for this test is: 05, find the following: A. Reject the null hypothesis B. There is insufficient evidence to reject the null hypothesis C. None of the...
im so grateful for all your help! this has helped me in so
many ways especially with all the things going on right now. thank
you so much !!
1
2
3
4
(1 point) 35 people are randomly selected and the accuracy of their wristwatches is checked, with positive errors representing watches that are ahead of the correct time and negative errors representing watches that are behind the correct time. The 35 values have a mean of 94sec and...
THANK YOU IN ADVANCE FOR HELPING ME! i appreciate this soo
much! this has helped me a lot especially throught times like this.
im so grateful for
your help
1
2
3
4
(1 point) The time needed for college students to complete a certain paper-and-pencil maze follows a normal distribution with a mean of 30 seconds and a standard deviation of 2.3 seconds. You wish to see if the mean time is changed by vigorous exercise, so you have...
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