Confidence Intervals for Mean
QUESTION 1 of 10
80% confidence level = 1.282 Stat 2
that is the answer that the book gave. Where did the extra 2 come from because the answer i got is 1.28
QUESTION 2 of 10
Hi! I am doing number 48 and for some reason I don't get the answer posted on the hw, i get 1.07, 5.13. However, when I do it on my graphing calculator like you did in the video, I get the right answer. I was wondering if you had a video of you doing the problem w/o the graphing calculator so I could see where I am doing it wrong.
QUESTION 3 of 10
Hello professor, I am having difficulties answering questions 9, 10, & 11 from the homework. I missed one class the first Friday after classes started so must likely you talked about it then. I apologize and thank for your help.
QUESTION 4 of 10
As I am finishing up studying, I had a question on 7.3. I reviewed when confidence level goes up the interval width increases and vise versa and that when sample size gets larger the confidence width gets smaller. Also, what can cause margin of error to increase and decrease. But I wanted to know on Now You Try It! For section 7.3 number 4 I am still a little confused about why increasing the confidence level requires a larger sample size?
The exact question was: What effect does choosing a higher confidence level have on the required sample size needed to estimate a mean within a given margin of error and having a given standard
Answer: . Increasing the confidence level will increase the required sample size.
Thank you for all your help this semester,
QUESTION 5 of 10
hi professor for questions 20 and 22 on 7.2 i am not getting the correct answer. when i divide 98% by 2 i get 0.49 so on the z-table i get 2.33. So when i do my calculations for 20 i get [821,420.062, 4,224,149.938] i did 2.33 x 4,065579/ square root of 31) and in 22 i got [2,549,528.835 , 6773113.165] i did 2.33 x (5046,335.26/ square root 31) then subtracted from the mean.
Can you catch where i am making a mistake? thank you
QUESTION 6 of 10
I got a similar answer for #20 on HW 7.2 but it wasn't as exact.
Can you help me find out what I did wrong?
My final answer is: [821,420.1 , 4,224,149.9]
My margin of error concluded to be: 1,701,364.938
My z alpha/2 is: 2.33
QUESTION 7 of 10
For homework problem 13 in section 7.2 if I cant find it on the t-table, I tried looking on the z-table but its giving me the wrong answer. Can you please go over how to do it.
13: The confidence level is 92%. The sample size is 49. The sample mean is 512, and the population
standard deviation is 28.
QUESTION 8 of 10
For HW Section 7.2 Number 12 I am confused when finding the alpha. Before I had just divided the confidence level by 2 but then watching the video for 7.1 you say to subtract the confidence level from 1 and then divide by half then look it up on the t-table. But when I multiply that by the standard deviation divided by the square root of the sample mean I dont get the right answer.
12. The confidence level is 99%. The sample size is 37. The sample mean is 25, and the population
standard deviation is 4.
QUESTION 9 of 10
I had a question on 7.2 Number 20:
20. A survey of thirty-one, 2005 Major League Baseball salaries for pitchers playing in the National
League had a mean of $2,522,785 and a standard deviation of $4,065,579. Construct the 98%
confidence interval for the true average salary for all NLMLB pitchers in 2005.
I was slightly off, I got (2655766.8, 238903.2) I tried to review it, but my answer was still wrong The same thing happened when I did the other similar problems about the baseball player.
QUESTION 10 of 10
For Homework 7.4 For number 42 n=128 so n-1=127 that number isnt on the t-table so how would you solve it? Also for homework 7.4 if the sample size is 30 or greater do we use the z-table?