Hw#8 Solution

8.18

x is N(69, 2.5)

a) N(69,2.5/3) →N(69,0.833)

b) N(69, 2.5/10)->N(69,0.25)

c) The sampling distribution in (b) has smaller variance.

8.28

a) X is U(8,16); P(x>13) = 3/8

b) N(12, 2.3/(square root of (50))), that is, N(12, 0.325)

P(x>13) = p(z>((13-12)/0.325)=0.0011, or

Normal cdf (13, E99, 12, 0.325) = 0.00105.

c) We cannot guarantee normality with n=10, n at least should be 30.

8.36

≈ N(0.42,sqrt(0.42(0.58)/150)), that is

N(0.42, 0.0403)

P(.035< <0.50) = p(0.35-0.42)/0.0403

<Z<(0.50-0.42)/0.0403)

=P(-1.74<z<1.98)=0.9761-0.0409=0.9352.

or use normalcdf(0.35,0.50,0.42,0.0403)=0.9352

8.38

The answer is Histogram A. The histogram should look approximately normal with a mean the same as the population mean, 60,000. This is portrayed in Histogram A.

8.40

X is N(4.5,0.6/6) that is N(4.5, 0.1)

a) p(x≤4.35) = p (z≤(4.35-4.5)/0.1) = p(z≤

-1.5)=0.0668.

or use normalcdf(-E99, 4.35, 4.5,0.1)=0.0668.

b) The CLT, central limit theorem.

9.2

a) true

b) false

c) true

d) false

9.8

a) Ho: p = 0.05 versus H1: p>0.05

b) The sample size is n=2000 and our estimate of p based on the sample is

^

p =/2000=0.0625. The p value can be found using 1-PropZTest with the calculator or by finding the probability

p(Z>=(0.0625-0.05)/(sqrt(0.05*(1-0.05)/2000))) = p(z>=2.565) = 0.00516

9.14

a) we should disagree. The correct interpretation is that when the null hypothesis is true, that is, when p=0.50, the chance of observing a z-value of 0.44 or larger is 33%

b) The p value would be (0.33*2)=0.66