business statistics

Sample 2994 Another sample of 100 different investors from the population of 200,000 people

Using the sample the estimated difference in proportions is 0.1-0.2=-0.1 So the zscore for that estimate is (-0-0.1)/0.0743=-2.68 Using wolfram alpha website P(Z<zscore) = P(Z<-2.68) =0.00368 So when you compare sample 2994 to the 4000 other samples you predict the rank to be 0.00368*4000=15

Theory you do not have to discuss this in the assignment

p1-p2 =0.1003 is positive

The 368 out of 4000 of the sample Estimates of p1-p2 are negative (less than 0) These samples are giving the opposite conclusion that the population , In the population the high risk investments have a higher probability of losing money however 368 of the sample make you give the opposite conclusion.

The proportion of samples that give the wrong conclusion is 368/4000=0.092

Using properties of the normal distribution , you can get a similar (but better answer) the estimates have a mean of 0.1 and standard deviation 0.0743 so an estimate of 0 has a zscore of (0-0.1)/0.0743=-1.34

P(Z<zscore) = P(Z<-1.34) = 0.09

In other words the estimate of 0 is the 9th percentile, this was found out using zscore and the normal distribution .

Some summaries of dataset used in Section 3 of the computer assignment (including zscores) ( the population and some samples )

A population of the returns of 200,000 investors

Note that return is express as a decimal, so a return of 0.05 is a return of 5%

So the difference in population means µ1- µ2=0.035-0.606=-0.256

There are 2000 samples so you can find 2000 estimates of µ1- µ2 =-0.0256

The average of the 2000 estimates of µ1- µ2 =-0.0256 The standard deviation of the 2000 estimates µ1- µ2 =0.0173

Sample 700 One sample of the returns of 100 investors from the population of 200,000 investors

So the difference in the sample means =0.0345-0.085=-0.0505 The zscore of this estimate =(-0.505- -0.0256)/ 0.0173=-1.44 using wolframalpha.com P(Z<zscore) = P(Z<-1.44)=0.075 So if you compare sample 700 to the 2000 other samples you expect it to have rank 2000*0.075=150

Sample 701 Another sample a different sample of 100 investors from the population of 200,000 investors

So the difference in the sample means =0.035-0.0618=-0.0268 The zscore of this estimate =(-0.0268- -0.0256)/ 0.0173=-0.07 using wolframalpha.com P(Z<zscore) = P(Z<-0.07)=0.471 So if you compare sample 701 to the 2000 other samples you expect it to have rank 2000*0.471=942

Sample757 Another sample a different sample of 100 investors from the population of 200,000 investors

So the difference in the sample means =0.0355-0.0256=0.0099 The zscore of this estimate =(0.0099 - -0.0256)/ 0.0173=2.05 using wolframalpha.com P(Z<zscore) = P(Z<2.05)=0.98 So if you compare sample 757 to the 2000 other samples you expect it to have rank 2000*0.98=1960

Sample244 Another sample a different sample of 100 investors from the population of 200,000 investors

So the difference in the sample means =0.0352-0.0352=0 The zscore of this estimate =(0 - -0.0256)/ 0.0173=1.48 using wolframalpha.com P(Z<zscore) = P(Z<1.48)=0.93 So if you compare sample 244 to the 2000 other samples you expect it to have rank 2000*0.93=1860

Theory you do not have to discuss this in the assignment

µ1- µ2  =0.1 is negative

The 136 out of 2000 of the sample Estimates of µ1- µ2  are positive (less than 0) These samples are giving the opposite conclusion that the population , based on the population the risky investments have a higher average but some samples say it has a lower average

The proportion of samples that give the wrong conclusion is 136/2000=0.068

Using properties of the normal distribution , you can get a similar (but better answer) the estimates have a mean of 0.1 and standard deviation 0.0743 so an estimate of 0 has a zscore of (0-0.1)/0.0743=1.48

P(Z<zscore) = P(Z<1.48) = 0.93 so 93% of the estimates will be below 0 so the proportion and give the correct answer but 7%=0.07 will give the wrong answer.

In other words the estimate of 0 is the 93rd percentile, this was found out using zscore and the normal distribution

Some summaries of dataset used in Section 4 of the computer assignment ,including zscores ( the population and some samples )

A population of 200,000 people is asked if they support a proposed change to the Business

population proportion of people that say yes is p=120,000/200,000=0.6

Since the sample has been split into 1000 samples there are 1000 sample estimates The average of the 1000 sample proportions is 0.6 The standard deviation of the 1000 sample proportions is 0.0357

Sample 700, A sample of 185 people are asked if they support the change. (the sample is from the population of 200,000 given above)

Sample proportion of people that say yes is =99/185 =0.5351

So the zscore =(0.5351-0.6)/0.0357=-1.82 P(Z<zscore)=P(Z<-1.82)=0.0344 So if we compare sample 700 to the 1000 other estimates then we expect the rank to be 0.0344*1000=0.0344=34

Sample 701, sample of a 190 people are asked if they support the change So a different group of people (a slightly bigger sample of 190) are selected from the population of 200,000

Sample proportion of people that say yes is = 119/190 =0.6263

So the zscore =(0.6263-0.6)/0.0357=0.74 P(Z<zscore)=P(Z<0.74)=0.7704 So if we compare sample 700 to the 1000 other estimates then we expect the rank to be 0.7704*1000=0.7704