Question

A random sample of 150 students is chosen from a population of 5000 students. If the mean IQ in the sample is 110 with a standard deviation of 10 what is the 95% confidence interval for the students mean IQ score?

Answer 1

@ash2326

Answer 2

@tkhunny

Answer 3

what do you know and what do you need?

Answer 4

I know the standard deviation, sample size, overall population, the mean IQ. I need to know the confidence interval for 95%

Answer 5

This is a straightforward problem once you've mastered the basics of confidence intervals. First off, you'll need the general formula, which looks like this:\[x-\bar \pm (z-critical-value) \frac{ sample~standard~deviation }{ \sqrt{n} }\].

By 'x -' I mean x-bar, the sample mean. (That's given in this problem.)
The z-critical value is actually a z-score, and its value depends upon the level of confidence desired. (In this problem, the z-crit-value is given.)
The sample standard deviation is given.
The variable n represents the number of samples. (Given).

What further help do you need?

Answer 6

I did a quick Internet search for "confidence interval" and found the following, among other probably helpful results:

http://stattrek.com/estimation/confidence-interval.aspx

Answer 7

so what is a confidence interval?

Answer 8

@nincompoop: If you'd like to know, please check out the link I provided (above). Glad you're curious about this.

Answer 9

So, it would theoretically turn out to be
\[5000\pm(.95)\frac{ 10 }{ 110 }\]

Answer 10

right?

Answer 11

I have a feeling I mixed a few of those up. My bad

Answer 12

Where'd that 5000 come from? That .95? The 0.10 is correct as the sample standard deviation. That 110 is not correct. Suggest that you review the info I've already provided AND look up the web site whose URL I've also provided. Let's not guess, OK?

Answer 13

The 5000 was the overall population
The .95 was the percentage
I hadn't looked at the website- I had been typing. I promise I wasn't trying to guess. I'm honestly really bad at this, and couldn't remember what all the terms meant. I'm sorry!

Answer 14

Though I had looked at your formula you gave me

Answer 15

I know what it is, but I dont think the person that posted the problem has any idea. just throwing out formula is very counter-intuitive specially when dealing with statistics, which deals with every bit of variation.

Answer 16

Based on your site, do we need to find the margin of error?

Answer 17

a formal introduction to the concept is very important in statistics, because every bit of information is crucial

I would suggest to do a bit of reading, then IF you still need help, seek out
http://onlinestatbook.com/2/estimation/mean.html

Answer 18

I've given you preliminary definitions of all of the quantities mentioned, and you can find more detailed definitions in that web site. Please stop saying, "I'm honestly really bad" and "I couldn't remember what all the terms meant." You'll improve in proportion to the amount of time and effort you put into learning this. Learning the definitions is essential.

Answer 19

Ok. May I try again? I won't try to put it in the equation without you checking it this time, so it will be a bit quicker.

Answer 20

A confidence interval is constructed for the purpose of ESTIMATING to a certain level of confidence what the population parameter is; in this case the population parameter is the MEAN VALUE of the population's IQ score. You will end up with an INTERVAL whose center is the sample mean (110), e. g., 110 plus or minus 3.5 points; this would indicate that "based upon the given sample data, the true mean IQ score for this population is (110-3.5, 110+3.5) at the 95% level of confidence."
You can of course do whatever you like, but to truly understand this problem, you'll need to know the vocabulary, be able to choose the proper quantities to substitute into the formula, and be able to explain the significance of your final result.

Answer 21

So...what is the sample mean?

Answer 22

The sample mean is 110

Answer 23

Or- I'm not quite sure- it may me 110-3.5 to 110+3.5

Answer 24

106.5-113.5

Answer 25

the sample mean is 110. Right.

What is the level of confidence involved here?

Please note: that 3.5 I gave you was STRICTLY AN EXAMPLE and does NOT apply to the problem you've posted.

Answer 26

That makes more sense- I was trying to figure that out. The level of confidence is 95%- Does that mean I take 95% of 110, then add and subtract it from 110?

Answer 27

Oops! 5%

Answer 28

Not 95%

Answer 29

That would be 104.5-115.5

Answer 30

Does your latest post follow the formula I've given you? does that formula call for multiplying 110 by 95% or by 5%?\[xbar \pm (z-critical-value) \frac{ sample~standard~deviation }{ \sqrt{n} }\]

Answer 31

I was trying to find the z-value

Answer 32

In a word: NO.
You are told that the level of confidence is 95%. The z-critical value associated with that level of confidence is 1.96. Have you ever seen that before?

Answer 33

I'm sorry- from now on I promise that I'll go bit by bit exactly as you tell me.

Answer 34

Yes, I have- Did you use a z-chart?

Answer 35

Guessing rarely helps.
You can find "z critical values" in several different ways; you can use a TI-84Plus calculator (if you have one); or you could refer to a table of z-critical values. One example of such a table is the large table labeled "Z-Table" on this page:

https://www.google.com/search?q=z-critical+value&rlz=1C1CHFX_enUS461US461&espv=2&tbm=isch&imgil=X82cUX-PmVskSM%253A%253BtUSGpcmmFdzumM%253Bhttp%25253A%25252F%25252F216.171.160.73%25252Fiapp%25252Fcritical-z-values&source=iu&usg=__H7zsJWP2fjjTr1_W6LF_LxS2z-A%3D&sa=X&ei=mYbvU4PRJqm68QHBm4DACw&ved=0CDsQ9QEwBQ&biw=1272&bih=683#facrc=_&imgdii=_&imgrc=tLwHQnbBL_bryM%253A%3BzJKTQJFRqYlmCM%3Bhttp%253A%252F%252Frmower.com%252Fstatistics%252Fprog_calc_stat%252FHow_to%252FzzShort_Z-Table_for_Critical_Numbers.jpg%3Bhttp%253A%252F%252Frmower.com%252Fstatistics%252FStat_HW%252F0804HW_sol.htm%3B622%3B583

Answer 36

If you look that up, and if you look in the "Z-Table" chart next to "95%", you'll see that the pertinent z-critical value is 1.96.

Answer 37

Thank you! I do see that.

Answer 38

Summarizing what we know:\[xbar \pm (z-critical-value) \frac{ \sigma }{ \sqrt{n} }\]

... we know that xbar is the same mean, 110; the pertinent z critical value is 1.96, and n, the number of samples, is 150. Let's say for now that sigma is 10. Then you could calculate a preliminary confidence interval.

Answer 39

Now we're looking at \[110\pm 1.96\frac{ 10 }{ \sqrt{150} }\]

Answer 40

Warning: this is not precisely the result you want, but I'd like for you to calculate this, to get an idea of what the correct result will look like.

Answer 41

Ok! Just one second-

Answer 42

Do you want the answer to be in decimal form, or have a radical?

Answer 43

decimal form, please

Answer 44

\[\approx1.6\]

Answer 45

without the 110

Answer 46

With:
108.4-111.6

Answer 47

Remember, you are aiming to come up with a CONFIDENCE INTERVAL.
you begin with the center, 110, and then subtract the "margin of error' from it; then you begin with 110 again, and add the "margin of error" to it.
How did you calculate your "margin of error?" type out the numbers you used.

Answer 48

\[1.96(\frac{ 10 }{ 12.24744871 })\]

Answer 49

Very good.
You 've written "108.4-111.6"; you mean a 'range' of 108.4 to 111.6.
Properly written, this should be (108.4, 111.6).

Answer 50

This is the basic procedure for constr4ucting a confidence interval.

Answer 51

I did mean a range- Thank you for the correction!

Answer 52

Now for the bad news. We have to re-work part of this problem to obtain the correct answer. Why? Because we use z-scores only if the population standard deviation, sigma, is known. Here, we don't know sigma; instead, we are told the the SAMPLE std. dev. is 10, right? Go back and check and convince yourself of that.

Answer 53

Oi

Answer 54

But I see

Answer 55

so we are given the SAMPLE std. dev., not sigma, the pop'n std dev

So, instead of using the table of z-scores, we have to use a different table, one named "t critical values." do you have such a table in fron t of you? Do you have a texbook that displays such a table? is such a table to be found in your online learning materials?

Answer 56

No, I don't that I know of. I'll try to find one online- just a moment please

Answer 57

https://www.google.com/search?q=t-critical+table&tbm=isch&imgil=8WkRBsfAjS6rgM%253A%253BZwHEiJnH1yHsbM%253Bhttp%25253A%25252F%25252Fwww.ruf.rice.edu%25252F~bioslabs%25252Ftools%25252Fstats%25252Fttable.html&source=iu&usg=__HprScqaS8aLGN9hUhO1ab2IAYGo%3D&sa=X&ei=t43vU_L5IaLksATLlICwCw&ved=0CCEQ9QEwAQ&biw=1366&bih=633#facrc=_&imgdii=_&imgrc=8WkRBsfAjS6rgM%253A%3BZwHEiJnH1yHsbM%3Bhttp%253A%252F%252Fwww.ruf.rice.edu%252F~bioslabs%252Ftools%252Fstats%252Fttable.gif%3Bhttp%253A%252F%252Fwww.ruf.rice.edu%252F~bioslabs%252Ftools%252Fstats%252Fttable.html%3B427%3B657

Answer 58

thanks for doing this look-up ... already your attitude is showing much improvement!

Answer 59

Have you heard of "degrees of freedom?"

Answer 60

Unfortunately, no

Answer 61

another concept we really do have to know. Basically, "df" (degrees of freedom) is defined as n-1. n is the sample size. Here, n=150. Thus, in this problem, df = ??

Answer 62

149?

Answer 63

Right. To find the correct "t-critical value" (not "z-critical value"), you look at such a table; on the left margin you'll usually find "df", which in this case is 149. At the top of the table there are numerals which correspond to the specified level of confidence, which is 95%.

Answer 64

I'm trying to find a different table- the one I found doesn't go up to 149

Answer 65

Here's the part which is a bit abstract (more for you because this is your first time): A 95% confidence interval signifies that your "alpha" is 1.00 - 0.95, or 0.05.

Answer 66

Don't bother finding a different table. The whole idea is that as df increases, the t critical value approaches the z critical value.

Answer 67

Please look at this excerpt from one of the t critical value tables we've found:

Answer 68

All I notice is that at the very bottom, infinity has a value at 5% of 1.960

Answer 69

Look at the column that is labeled 5% (from 100%-95%). Move your focus downward to the very bottom; you'll see first that df =120, and then df = infinity. Note that if df=120, the t-critical value is 1.98, whereas if df goes to infiinity, the t crritical value is 1.96, the same as the z-critical value we found earlier (1.96). ok WITH THIS?

Answer 70

yOU'RE ABsolutely right.

Answer 71

In other words, our check is correct and was worth the time, but our t-critical value is 1.96, just as the 1.96 z-critical value we used earlier.
This means that your confidence interval is correct as it stands.

Answer 72

Awesome! Does that mean that my previous range was correct? (The one with the correct margin of error of 1.6)

Answer 73

Note that if df were 16, the t-critical value would be greaster: 2.120.
But as df increases towards infinity, the t critical value takes on the value of the z-critical value. Yes, your previous confidence interval is correct as is.

Answer 74

Thank you so much for sticking with me through this problem! Thank you for explaining every bit of the way, so I could actually learn the material!

Answer 75

that's what I'm here for. You contributed by becoming much more willing to spend the time and effort necessary to learn and undrstanding this material. Thank YOU.

Answer 76

Yay! Only 13 problems left after that!

Answer 77

Yay....ONLY 13!
However, the work we've done this morning should enable you to go thru the other problems faster.
Although I need to get off the 'Net now, I'll probably be on OpenStudy again later. I urge you to do as much as you can by yourself on the remaining problems and then (only then)_ ask for help.

Answer 78

Ok! Thank you for the extensive help!

Answer 79

My pleasure! Best of luck; hope to work with you again later on.