Sohh Who Here Took Statistics

Ronald McDonald

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I'm stuck on these last 5 problems :martin:

I'm a few % points from a C. Help me out:

1. The following # are the ages of randomly selected Race Car Drivers. Construct a 98% confidence interval estimate of the mean age of all the race car drivers

32 32 33 33 41 29 38 32 33 23 27 45 52 29 25
 

Ronald McDonald

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2. The Wechsler IQ Test is designed so that the mean is 100 and the standard deviation is 15. Find the sample size necessary to estimate the mean IQ score of professional pilots. We want to be 90% confident that our sample mean is within 3 IQ points of the true mean. The mean for this population is clearly greater than 100. The standard deviation for this population is probably less than 15 because it is a group with less variation than a group randomly selected from the general population; therefore, if we use standard deviation = 15, we are being conservative by usin value that will make the sample size at least as large as necessary. Assume then that standard deviation = 15 and determine the required sample size. Does the sample size appear to be reasonable?

^This shyt here :martin:
 

Ronald McDonald

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3. You want to estimate the mean SAT score of all college applicants. First use the range rule of thumb to make a rough estimate of the standard deviation of these scores. Possible SAT scores range from 600 to 2400. Use the estimated standard deviation to determine the sample size corresponding to a 98% confidence level and a margin of error of 100 points. What isnt quite right with this exercise?

The standard deviation is 450. The answer is 110. I dont know the work used to get it though :jbhmm:
 
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Snoopy Loops

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95% confidence interval

Calculate the mean
Calculate the standard deviation(std)
95% Confidence interval= mean +/- (2.776*std/sqrt (n))

The 2.776 is from the t test table.
 

Snoopy Loops

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I like your dedication :obama:

So I got the 2nd one with help from youtube. I'll put the link below.

So Known variables are: population mean = 100 , standard dev (std) = 15 , margin of error(M.E) = 3, alpha value =90% (0.9)
Unknowns: sample size and Ttest( because you don't have sample size and hence the degrees of freedom(df=n-1))

M.E = Ttest * std/sqrt(n)

The thing here is you cant use the Ttest because you dont have the sample size. Its a necessary value to find the T value, if you check the table, you'll the degrees of freedom which is n-1.

So you resort to the Z value from the tables (0.9 alpha value)which is 1.6443.

Plug 1.6443 into the M.E formula.
find n

n = (1.6443 * 15/3)^2
n=67.5=68 (not sure if you round up or down tho)

Here's the link
 

Snoopy Loops

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For the last one, just follow the video. It's the same thing.

Check out youtube vids for all this kinda stuff.

Also you might check with someone else if they used the same methods.
 
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I hate when some jobs in my field (about 10-15%) are like you should be good at STAT and SPSS and shyt.

I'm like listen I calculate statistical significance and confidence off experience and instinct. fukk all those formulas.

My mind can do statistics unconsciously, but when it comes to making actual formulas it feels very awkward and unnatural.

There are actual business decisions based off of data analysis using statistical modeling. The thing is it depends on the industry.

If you are a biochemical engineer then yes you will certainly want to use formulas and regimented proven applied principles to avoid blowing the lab up and killing everyone.

But if you're talking about online customer acquisition, behavioral analysis of purchasing patterns honestly this stat shyt is over-rated. We are evaluating our own behaviors as objectively as possible - But at the end of the day do we really need a linear regression model to know if our business plan will be profitable?

Probably not.
 

TL15

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Gimme a sec breh....I'll be back

Rqhfvi.gif
 

Ronald McDonald

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Last two:

5. sample size = 40
df = 39
Mean = 172

Cosntruct a 95% confidence interval estimate of the population mean

6. You have been given the task of estimating the percentage of Southwest flights that arrive on time, which is no later than 15 minutes after the scheduled arrival time. How many flights must you survey in order to be 80% confident that your estimate is within three percentage points of the true population percentage?

a. Assume that nothing is known about the percentage of on-time southwest Flights
b. Assume that for a recent year, 84 of Southwest flights were on time
c. Given that the sample size is relatively small, can you select Southwest flights between New York and San Fransisco?
 

tater

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I took this maybe two years ago, but I don't remember half this stuff lol. Hope I never need it :lupe:.

But yahoo answers always comes through if I just need an answer to a problem. You can post anonymously too.
 
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