Statistics and probability practice problems – Term Paper

Statistics problem sets

Section A

Question A1

a)X is a hyper-geometric distribution since it is always associated with selection without replacement.

b)Formula: P(X=x) is given by f(x) = 

Procedure

Because selection is without replacement, the Probability Mass Function needs to be 

N1        N2

x          n-x

Such that; N1 is the items of the 1st type         N1+N2

                  N2 is the item of the 2nd type          n

                   X is the non-negative integer satisfying the condition x <= n, x<= N1, n-x <= N2

                   N=N1+N2

Question A 2

a)The probability that the mean of the sample is greater than 150 is calculated using the formula.  

Procedure 

Substitute the respective values in the formula from the information given such that:  

 =    = 1.25

 From the Z score table the P(Z > 1.25) is given as 0.8944. therefore the probability that the mean of the sample is greater than 150 is 0.8944. 

b)Find the minimum sample size that can make the probability in  A2(a) equal to 0.001

Procedure 

Find the value that corresponds with the value 0.001 from the Z score table. Which is 3.10.  

Then make the value found the answer of the function in A2(a), the evaluate such that :  = 3.10

Then find n such that :

 = 3.1

2 = 

 = 24.8

n = 615

Therefore, the minimum sample size that can make the probability correspond 0.001 is 615.

Probability problems

Question A 3

a)The simple linear regression model is given by

Y = a + bx + note the error term is negligible 

Y is the response variable or the dependent variable  

X is the predictor or independent variable 

a is the slope/gradient/constant

When y changes by 1 unit x changes by b units if the a units are held constant. 

b)The least square estimator minimizes the squared errors between the expected and the observed values of the parameters α and β in the process of fitting a linear regression model to data.

c)A relationship exist between least square estimator and maximum likelihood estimator during the estimation of the parameters α and β. This is because both methods yield the same results.

Question A4 

a)To prove that f(x) is a probability function we need to show that the integration of f(x) is equal to 1.

Procedure 

   b)Derive the probability P(x<t)

Procedure 

C procedure

Section B

Question B1

a)(i) The null hypothesis is H0: = 350 grams 

     The alternative hypothesis is H1: 350 grams    

(ii) The test statistic for this population is t at (n-1) degrees of freedom

The formula used to calculate test statistic is  

From the information provided 

S = 20

= 362

= 350

n = 9

Therefore

t(n-1)= 

t(n-1) = 1.8

 t(8) = 1.8

(iii) Since t at t(n-1)  is greater than ttab which is 1.397 we therefore reject the null hypothesis H0 and accept the H1 an conclude that is not equal to 350.

b)(i) The null hypothesis H0: = 350 grams 

     The alternative hypothesis H1:  > 350 grams

(ii) The test statistic for this distribution is which is calculated as follows.

From the data provided 

  = 350

x = 327, 358, 374, 359, 397, 367, 331, 368, 385

Therefore =  

= 362

c)The standard deviation has changed since most of the data except 350 and 359 are lying further away from the mean. Thus, it is reasonable to suggest that standard deviation has changed. Moreover, the deviation from the mean is evidence of variance in the data thus changes in the standard deviation. 

At α = 0.05 t(9-1) = t(8, 0.05) = 1.86.

Therefore, since ttab 1.86 is greater than tcalc 1.5532 we reject the null hypothesis and conclude that indeed <= 350 grams. Moreover, a significant level of 0.05 increases the precision. 

Problem on probability

Question B2

Least square estimate

By method of moments

Maximum likelihood estimate

Consistent estimator for poisson

Question B3

a)

b)(i) this distribution is normal with mean  and standard deviation 

(ii) This distribution is normal with mean  but with a much smaller standard deviation of  this is a distribution of the sample mean   

Question B4

a)In this case, there are 10! Likely arrangements based on the sitting arrangement. Moreover, there are a number of combinations possible in this case. To count them we consider nine objects among them Alan, Barbara sitting together, and the remaining children make the other combination. The total number of possible combination is 9! The total number of arrangement that can have Alan and Barbara together is 2*9! Thus the required probability is 

Therefore the probability of Alan and Barbara siting together is 

b)In the case of Clare, Daniel, and Edward sitting together, we consider 8 objects. There are 3! Possible sitting arrangement for 3 children in the group and 8 ways of combinations. Thus the required probability is 

c)In this case, we already know the probability of Clare, Daniel, and Edward siting together. Thus in order to find the probability of Clare, Daniel, and Edward sitting together, Alan, and Barbara siting apart, we can subtract the probabilities. Such that

Procedure 

 (Here, we use AB to denote the event that A and B sit together and CDE the event that C, D and E sit together.) 

We find  by considering the 7! Possible combinations of seven objects, that is A and B, and C, D and E, and the five remaining children. Taking into account the 2 possible orders for A and B and the 3! Possible orders for C, D and E, we obtain 

Therefore the solution to the problem is 

d)Here we first consider the other seven children. They can have  possible combinations. Here there are 2 possible ways of arranging C, D and E in the remaining seats with D siting between C and E. hence the solution will be.