a) At 90% confidence level, the critical value is z_{0.05} = 1.645
The 90% confidence interval is
b) Since the interval does not contain 5.61, so the confidence interval contradict the claim.
Question 8 of 18 (1 point) Attempt 1 of Unlimited View question in a popur! 72...
Question 7 of 22 (1 point) Attempt 1 of Unlimited View question in a pour 82 Section Exercise 21 (cale) Online courses: A sample of 265 students who were taking online courses were asked to describe their overall impression of online learning on a scale of 1-7, with 7 representing the most favorable impression. The average score was 5.61, and the standard deviation was 0.94. (a) Construct a 99.5% confidence interval for the mean score. (b) Assume that the mean...
A sample of 250 students who were taking online courses were asked to describe their overall impression of online learning on a scale of 1-7 with 7 representing the most favorable impression. The average score was 4.98 and the standard deviation was 0.87. Construct a 90% confidence interval for the mean score. Round your final answers to 3 decimal places. The 90% confidence interval for the mean score is: ( , )
Question 4 of 18 (1 point) Attempt 2 of Unlimited View question. In a popuR! 7.1 Section Exercise 37-38 A sample of size n = 92 is drawn from a normal population whose standard deviation is a = 5.6. The sample mean is x = 46.06. Part 1 of 2 (a) Construct a 95% confidence interval for Jl. Round the answer to at least two decimal places. A 95% confidence interval for the mean is <u< . Part 2 of...
Question 12 of 18 (1 point) Attempt 1 of Unlimited View question in a popur! 7.3 Section Exercise 1 For the given confidence level and values of x and n, find the following. x=48, n = 97, confidence level 99.9% Part: 0/3 Part 1 of 3 (a) Find the point estimate. Round the answers to at least four decimal places, if necessary. The point estimate for the given data is
Question 1 of 13 (1 point) Attempt 1 of Unlimited View question in a popup 8.1 Section Exercise A sample of size n=74 is drawn from a normal population whose standard deviation is o=7.2. The sample mean is x =43.56. Part 1 of 2 (a) Construct a 99.8% confidence interval for u. Round the answer to at least two decimal places. A 99.8% confidence interval for the mean is <u< . Part 2 of 2 (b) If the population were...
Question 16 of 18 (1 point) Attempt 1 of Unlimited 74 Section Exercise 12 (cal Ages of students: A simple random sample of 110 U.S. college students had a mean age of 23.02 years. Assume the population standard deviation is o = 4.77 years. Construct a 98% confidence interval for the mean age of U.S. college students. Round the answers to two decimal places. A 98% confidence interval for the mean age of U.S. college students is
7.1 Section Exercise Question 3 of 18 (1 point) Altempt 3 of Unlimited View questlon in a popup 7 A sample of size n 87 is drawn from a population whose standard deviation is a Part 1 of 2 (a) Find the margin of error for a 90% confidence interval for . Round the answer to at least three decimal places. The margin of error for a 90% confidence interval for u is Part 2 of 2 (b) If the...
0_U-IgNsikasNWEDBA9PVVRe4tCL6xedC30WT75R50Xmai_KAADILAjqzxPmXWKWIKC78EfixucBBnZ8UYHNdNiOwplSM. Homework 14 Question 8 of 11 (1 point) Question Attempt 2 of Unlimited ✓2 ✓3 ✓5 6 ✓7 8 ✓9 ✓ 10 ✓ 11 A psychologist is interested in the mean IQ score of a given group of children. It is known that the IQ scores of the group have a standard deviation of 13. The psychologist randomly selects 100 children from this group and finds that their mean IQ score is 103. Based on this sample, find a...
Homework 12 Question 5 of 5 (1 point) Question Attempt 1 of Unlimited E The mean SAT score in mathematics, H, is 544. The standard deviation of these scores is 41. A special preparation course claims that its graduates will score higher, on average, than the mean score 544. A random sample of 90 students completed the course, and their mean SAT score in mathematics was 552. At the 0.01 level of significance, can we conclude that the preparation course...
stion 14 of 18 (1 point) Attempt 1 of Unlimited View question in a popun iart phone: Among 249 cell phone owners aged 18-24 surveyed, 108 said their phone was an Android phone. rform the following. Part 1 of 3 (a) Find a point estimate for the proportion of cell phone owners aged 18-24 who have an Android phone. Round the answer to at least three decimal places. The point estimate for the proportion of cell phone owners aged 18...