STATISTICS

Laboratory Assignment 2

Laboratory Assignment 2

Answer No. 2

A = spatial skills score

B = verbal skills score

Thus,

A = N (63, 8^ 2)

B = N (71, 5^ 2)

a) The percentage of students who score above 76 in the verbal skills test

= 100 x p (B > 76.5)

= 100 x p (Z > (76.5 - 71)/ 5); where Z is N (0, 1^ 2), the standard Normal variant

= 100 x p (Z > 1.1)

= 100 x (1 - F (1.1))

= 100 x (1 - 0.864) table value

= 13.6 %

b) If a person scores 71 on the spatial skills test and 73 on the verbal skills then we first need to find their standardized scores

Z 1 = (71 - 63)/ 8 = 1

Z 2 = (73 - 71)/ 5 = 0.4

t- scores are calculated using the formula that is 10(z-value) + 50; thus,

t1 = 10 x 1+50= 60

t 2 = 10 x 0.4+50= 54

It can be said that the individual's score is better for spatial skills

If X = A+B then X is N (63+71, 8²+5²) = N (134, 89)

c) The mean of the sum of scores on the two tests = 63+71= 134

d) The SD of the total scores on the two tests is v(8²+5²) = v89 ˜ 9.43

e) The probability that a person obtains a total score of at least 140 on the two tests

= p (X > 139.5)

= p (Z > (139.5 - 134)/ v89)

= p (Z > 0.583)

= 1 - 0.72 table value

= 0.28

Answer No. 3

If, A = number of under filled bottles, A ranges from 0 to 3 with frequency value of 11, 15, 11, 3

It is also assumed that the chances of a bottle being under filled is that is half, thus, it is a success; otherwise, it is a failure.

H0: Binomial offers an appropriate model.

Ha: It is not an appropriate model.

Moreover, if, n is 3, then;

X = 0, 1, 2 , 3

P (a=0) = 3C0 (0.5) ^ 0 (1 - 0.5) ^ 3 = 0.125

E (f) = (40) (0.125) = 5

P (x = 1) = 3C1 (0.5) ^ 1 (1 - 0.5) ^ 2 = 0.375

E (f) = (40) x (0.375) = 15

P (x = 2) = 3C2 (0.5) ^ 2 (1 - 0.5) ^ 1= 0.375

E (f) = (40) x (0.375)= 15

P(x = 3) = 3C3 (0.5) ^ 3 (1 - 0.5) ^ 0= 0.125

E (f) = (40) x (0.125)= 5

Chi-squared = (11 - 5) ^ 2 / 5 + (15 - 15) ^ 2 / 15 + (11 - 15) ^ 2 / 15 + (3 - 5) ^2 / 5 = 9.07

In comparison to critical chi-square with 3 df that is alpha level of 0.05, the value of degree of freedom is 7.82. Thus, H0 is rejected as chi-square exceeds critical chi-square value.

Answer No. 4

Sample size is small for conducing statistical analysis. The study test for a significant difference in the two proportions, that is p2 hat=2/ 6 and p1hat=3/ 16.

For a null hypothesis there is no difference the estimated ...