stats 2 final problems final solutions

8/12/2019 Stats 2 Final Problems Final Solutions

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Stats 2 - Review Problems

1) a) Two different midterm were given to stats sections A and B. Test the claim that the midterm given to section A was harderthan that given to section B. There were 35 students sampled in section A which had a class average of 55% and SD of 5%.There were 25 students in section B which had a class average of 59% and SD of 3%. Assume equal variances.

Note that if the midterm given to section A was harder, the score would be lower. We must do a 2-sample T-test becausethe two groups are independent. Note also that since the sample sizes are not the same, it cannot be dependent. Becausewe have a < sign in Ha, we are doing a 1-tailed test.

Ho: a= b Ha: a< b Ho: a- b=0 Ha: a- b 5 d = a- b df = n - 1 = 32-1=31 =0.05 d = 4 sd=1.5 n=32T-stat = ( )

= ( )

= -3.7714

Look up 3.7714 in T-table at 31 df and we find that it falls to the right of the last column corresponding to a probability of0.0005 one tailed. *This probability will depend on the table you have available.

Be careful, as the rejection are will be to the right of this line meaning p > 1-0.0005 => p >0.9995T-crit = 1.697 from T-table at 30 df. Again, we are dealing with the right hand side of the curve, so this remains positive.

CI =( ) - T-crit*( )= ( ) - 1.697*( ) = (3.5500,100)Do not reject Ho because p> , T-stat < T-crit, and our CI does contain d or "5".

c) If in parts a) and b), we increase the sample sizes to 105, how would your answers change? Show the new solution.


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*a) Note that if the sample size increases to over 100 we have a large sample and can thus assume that the sample SD isapproximately equal to the population SD. We then end up calculating a Z-stat instead of a T-stat as shown below. Noticethat for the large sample formula we cannot use a pooled standard deviation.

Ho: a= b Ha: a< b Ho: a- b=0 Ha: a- b | Z-crit|.

*b)

Ho: d= 5 Ha: d> 5 d = a- b =0.05 d = 4 d=1.5 n=105Z-stat = ( )

= ( )

= -6.8313

Look up -6.8313 in Z-table and we find a value of 0.5 *This probability will depend on the table you have available.

Be careful, as the rejection are will be to the right of this line meaning p = 0.5+0.5=1Z-crit = 1.645. Again, we are dealing with the right hand side of the curve, so this remains positive.Do not reject Ho because p> , Z-stat < Z-crit.

d) If in parts a) and b), there were only 200 students in each stats section, how would your answers change? Show the newsolution.

You would have to apply the FPCF, not covered for 2 sample. I will show how to apply with 1 sample.

e) Given the test you chose to use in part a), what boxplots(s) should you be looking at to determine normality? If the data wasnot normal, what would your hypothesis look like?

Because this is an independent test, we would want to look at the boxplot from each of the samples. If even one was notnormal we would perform a Mann Whitney test with the following Hypothesis:

Ho: M a= Mb Ha: M a< Mb or Ho: M a- Mb=0 Ha: M a- Mb 5 where, M d = Ma- Mb

2a) The profs for the class claim that exactly 75% of students should pass the final. To test this you sample 50 students fromlast year's class and find that 36 of them passed the final, test the profs claim.

We are now dealing with binary data, and we have 1 sample so we will do a 1-proportion test, but first we must validate theassumptions:


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Ho: p= 0.75 Ha: p 0.75 p hat = 36/50=0.74 n=50 np=37.5>10 nq=12.5>10Assumptions are met, 1-proportion test is ok. =0.05

Z-stat = ( ) ( ) =( ) ( ) = -0.1633

Look up -0.1633 in Z-table and we find a value of 0.4364. (This was using the negative table). Because we have a 2 tailed

test, we have a rejection area at both ends. So p-value=2*0.4364 = 0.8728 Do not reject Ho because p> .

b) The profs for the class also claim that less than 5% of students should score over 90 percent on the final exam. To test thisyou sample 50 students from last year's class and find that 2 scored over 90% , test the profs claim.

Ho: p= 0.05 Ha: p< 0.05 p hat = 2/50=0.04 n=50 np=210Assumptions are not met, we must use binomial.

We need to calculate our P-value as P(x p f + 0.1 Ho: p p - p f =0.1 Ha: p p - p f >0.1 =0.05 p p_hat = 15/20=0.75 n p=20 p f_hat = 12/20=0.6 n f =20Z-stat => apply formula in number 11 on the cheat sheetZ-stat = 0.3412

Look up 0.3412 in the Z-table and we find a value of 0.6331.p=1-0.6331 = 0.3669

Do not reject Ho because p> .

4) A local sports store assumes that they see twice as many customers on saturdays and sundays than they do during the rest ofthe weekdays. Test to see if this distribution of customers is accurate using last week's customer counts given below.

Customers Expected Chi-SquaredMonday 24 33 2.4545Tuesday 25 33 1.9394Wednesday 33 33 0.0000Thursday 15 33 9.8182Friday 44 33 3.6667Saturday 74 66 0.9697

Sunday 82 66 3.8788Total 297 297 22.7273

Step 1 - Calculate the number of observations by adding up all observed values.Step 2 - Derive Expected Values. Let x be the expected number of customers on a weekday. Let 2x be the number ofcustomers on a weekend. So the total weekly customers would be x + x + x + x+ x + 2x + 2x = 9x = 297. Solving for x, we get avalue of 33. So we would expect 33 customers on weekdays and 66 customers on weekends. Step 3 - Calculate chi 2 value for each cell. chi 2 = (o-e) 2/e. chi 2 (Monday) = (24-33) 2/ 33 = 2.4545Step 4 - Calculate Chi 2Stat value by adding up the chi

2 values from each cell.Chi2Stat = 2.4545 + 1.9394 + 0 + 9.8182 + 3.6667 + 0.9697 + 3.8788 = 22.7273


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Step 5 - Hypothesis Test - Ho: Data follows described distribution Ha: Data follows some other distributiondf=(k-1) = (7-1) = 6 Chi 2crit = 12.5916Since Chi 2Stat > Chi

2crit we reject Ho, data does not follow the described distribution.

P-value = less than 0.005

stats 2 final problems final solutions

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