CALCULATION PROBLEMS (10 points each) 1. At Plattsburgh Inc. 200 employees had c

Question

CALCULATION PROBLEMS (10 points each) 1. At Plattsburgh Inc. 200 employees had colds. 150 employees who did no exercising had remainders of the employees with colds were involved in a weekly ex ercise program. Half of the 1,000 employees burghnc. were involved in a weekly exercise program. Design a joint probability table and find the probability that a randomly selected employee: Exercise (E) No exercise (E Total 200 800 1000 Cold (C) 50 No cold (Cc)450 150 350 500 Total 500 a) didn't get cold b) got cold and was involved in a weekly exercise program c) didn't get cold and was not involved in a weekly exercise program d) Prove if "exercise" and "not getting cold" are statistically dependent or independent. 2. Battery manufacturers compete on the basis of the amount of time their products last in cameras and toys manufacturer of alkaline batteries has observed that its batteries last for an average of 26 hours when used in racing car. The amount of time is nomally distributed with a standard deviation of 2.5 hours. a) What is the probability that the battery lasts less than 28 hours? b) What is the probability that the battery lasts less than 25 hours? c) What is the probability that the battery lasts longer than 29 hours? d) What is the probability that the battery lasts longer than 24 hours?

Explanation / Answer

(1)

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(a) Probability that the employees didn't get cold = P(C') = 0.80

(b) Prob (got cold and involved in excercise) = P(CE) = 0.05

(c) Prob (didn't get cold and not involved in excercise) = P(C'E') = 0.35

(d)

A Chi-square test for independence will determine the independence of these two categories.

H0: the row and the column variables are independent
Ha: the row and the column variables are associated

The test statistic SumiSumj (Oij - Eij)2 / Eij = 62.5 which is greater than the critical value at a 'df' of (2-1)*(2-1) = 1 and confidence level of 95% which is 3.84. So, we reject the null hypothesis that the factors are independent and assert that they are associated (dependent)

(2)

Given than Life (X) ~ Normal (m=26, s=2.5)

Z = (X - m)/s where Z ~ Normal(0,1)

(a)

P(X < 28) = P((X - m)/s < (28 - 26)/2.5) = P(Z < 0.80) = 0.788 (from standard normal tables)

(b)

P(X < 25) = P((X - m)/s < (25 - 26)/2.5) = P(Z < -0.4) = P(Z > 0.4) (from symmetry) = 1 - P(Z < 0.4) = 0.345

(c)

P(X > 29) = P((X - m)/s > (29 - 26)/2.5) = P(Z > 1.2) = 1 - P(Z<1.2) = 0.115

(d)

P(X > 24) = P((X - ?)/m > (24 - 26)/2.5) = P(Z > -0.8) = P(Z < 0.8) (from symmetry) = 0.788

(3)

There are four samples (for four colors) and each sample has a sample size of 6.

(a)

H0: m1=m2=m3=m4 where 'mk' is the population mean number of insects for the i-th color, i=1,2,3,4
H1: At least one of m1, m2, m3, and m4 is different than the others

(b)

Since 'F' is greater than 'F crit' and also since the P-value is less than 0.05, we reject the null hypothesis at 95% confidence level and assert that the mean number of insects will not be all same for the four colors.

Exercise
(E) No Exercise
(E') Total Cold (C) 50 150 200 No Cold (C') 450 350 800 Total 500 500 Exercise
(E) No Exercise
(E') Total Cold (C) P(CE) = 50/1000 = 0.05 P(CE') =150/1000 = 0.15 P(C) = 0.25 No Cold (C') P(C'E) = 450/1000 = 0.45 P(C'E') = 350/1000 = 0.35 P(C') = 0.80 Total P(E) = 0.50 P(E') = 0.50 1.00

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