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Course Documents
Chapter 1 - Intro
Chapter 2 - Methods for Describing Sets of Data
Chapter 3 - Probability
Chapter 4 - Discrete Random Variables
Chapter 5 - Normal Random Variables
Chapter 6 - Sampling Distributions
Chapter 7 - Confidence Intervals
Chapter 8 - Tests of Hypothesis: One Sample
Chapter 9 - Confidence Intervals and Hypothesis Tests: Two Samples
Sample Exam I: Chapters 1 & 2
Sample Exam II: Chapters 3 & 4
Sample Exam III: Chapters 5 & 6
Sample Exam IV: Chapters 7 & 8
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
Posted to STATS 1 on Sunday, November 24, 2013 Replies: 3
Hi Natassia,
Your answer is a little off because of the critical z value that you used. You did not use the t-table to find the critical value. You used the z-table to find it. You should use the t-table for 99%, 98%, 95%, 90%, and 80% confidence.
I actually created a video for this problem to answer another student's question. Click the 'Ask the Professor' link and find the question category for 'Confidence Interval for Mean'.
Watch the video for more detail,
Professor McGuckian
Hi Natassia,
In that section, you are trying to distinguish between when to use a z-critical value or a t-critical value. Remember, we are using the simplified rule that states if n > 30, we will find a z-critical value, otherwise we will find a t-critical value.
In the problem you asked about, since n = 128, we need a z-critical value, so you do not need to find degrees of freedom to find the 1.282. All you need to do is to find alpha/2 and go to the t-table. The answer you need will appear in the last row of the table.
Hope that answers your question,
Professor McGuckian
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