Question

abswer all questions please!

10. Given the e variances of the two samples below, find the test value and the degrees of freedom that should be used in an Ftest. Sample 1 Sample 2 13 28 Variance Sample size A) test value 0.54; d.f.N. -27 and d.f.D.-13 B) test value 1.86; d.f.N. 27 and d.f.D.-13 C) test value 1.36; d.f.N. 28 and d.f.D. 14 D) test value 3.45; d.f.N.-28 and d.f.D.-14 11. The mean weight of loads of rock is 45.0 tons with a standard deviation of 12.0 tons. If 36 loads are chosen at random for a weight check, find the probability that the mean weight of those loads is less than 44.0 tons. Assume that the variable is normally distributed. A) 78.77 B) 71.23 C) 21.23 D) 28.77 12. How many different ways can a teacher select 5 students from a class of 15 students to perform a different classroom task? A) 75 B) 30,240 3,003 D) 360,360 15Cs 13. A random sample of magnesium concentrations (in parts per million, or ppm) in ground water from various locations follows. Estimate the mean concentration of magnesium in ppm with 90% confidence. Assume σ-5. 119 97 87 56 33 75 17 103 61 105 49 60 87 37 123 24 20 117 28 69 31 22 78 57 8 52 44 67 19 99 10 28 54 88 93 C) 56.3< u<64.7 D) 55.5 < < 65.5 59.8 < μ < 61.2 58.0< <63.0 A) B) 14. A scientist claims that only 66% of geese in his area fly south for the winter. He tags 65 random geese in the summer and finds that 19 of them do not fly south in the winter. a .05, is the scientist's belief warranted? A) No, because the test value-0.81 is in the noncritical region. B) Yes, because the test value -0.89 is in the noncritical region. C) Yes, because the test value 0.81 is in the noncritical region. If No, because the test value 0.89 is in the critical region. D) 15. The average age of doctors in a certain hospital is 42.0 years old. Suppose the distribution of ages is normal and has a standard deviation of 8.0 years. If 25 doctors are chosen at random for a committee, find the probability that the average age of those doctors is less than 42.6 years. Assume that the variable is normally distributed. A) 49.5% B) 64.8% C) 35.2% D) 14.8%

Answer 1

10)

test value =(13/7)=1.86 ; option B is correct

11)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	45
std deviation   =σ=	12.0000
sample size       =n=	36
std error=σx̅=σ/√n=	2.0000

probability =

P(X<44)

=

P(Z<-0.5)=

0.3085

option D 28.77 seems to be closest and correct

13) option C is correct

14)

option C is correc

15)

probability =

P(X<42.6)

=

P(Z<0.38)=

0.648~64.8%

option B is correct