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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

Professor Mcguckian
11/24/2013
6:56 PM EST

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

.
11/24/2013
10:26 PM EST
Thank you I understand now.
in 7.3, what do you do when the d.f value is not on the t-table for example 42. d.f= 127 but that is not of the t-table?

Professor Mcguckian
11/24/2013
11:56 PM EST

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.