Exercises 2.1 10

Exercise 2.1.

n  5,

p  0.25

0.25  0.75  pˆ   0.00375  0.06124 50 pˆ  0.25 Z 0.06124 0.27  0.25   0.23  0.25 P (0.23 p pˆ p 0.27)  P  pzp  0.06124 0.06124    P (0.326 p z p 0.326)  1  2  0.3707  0.2586

Exercise 2.2. First we calculate the standard devirative of the mean from the population standard deviation

 x $4, 000   $565.68 n 50 E  X    X    $55, 000 x 

We know that

Because the sample mean X is normally distributed, we can use to calculate

Z So

X  $56,500  $55, 000   2.65 $565.68 X / n

P  X  $56.500   P( Z  2.65)

 .5000  .4960  .0040 Thus there is .4 percent chance of avarage annual earnings being more than $56,500 in a group of 50 assisstant marketing professors.

Exercise 2.4.

s 140 x  z / 2   357.60  1.96   330.16;385.04  n 100

Exercise 2.5.

.65(1 − .65) .65(1 − .65) .65 − 1.96 < p < .65 + 1.96 30 30 .479 < p < .821 Exercise 2.6.

α=.05

n=30

2 χn2−1,α / 2 = χ29 ,.025 = 45.7222 2 χn2−1,1−α / 2 = χ29 ,.975 = 16.0471

⇒ 95% c.i.

( 29)(5.5) ( 29)(5.5) <σ 2 < 45.7222 16.0471 3.488 < σ 2 < 9.94

Exercise 2.7. The decision rule is to reject H0 in favor of H1 if

X1 − X 2 1

2 2 2

s s + n1 n2

> Zα

(attn: n1=n2=48>30)

In this example,

X 1 = .899, s12 = .247, n1 = 48, X 2 = .703, s 22 = .320, and n2 = 48, so X1 − X 2 2 1

2 2

s s + n1 n2

=

.899 − .703 .247 .320 + 48 48

= 1.80

From Table for Z statistic in Appendix, we find that the value of α corresponding to zα=1.80 is .0359. Hence the null hypothesis can be rejected at all levels of significance greater than 3.59 percent. If the null hypothesis of equality of population means were true, the probability of observing a sample result as extreme as or more extreme than that found would be .0359. This is quite strong evidence against the null hypothesis of equality of these means, suggesting rather a decrease in the mean risk premium after the “Franklin Message”.

Exercise 2.8. The decision rule is to reject H0 in favor of H1 if X 1 −X 2 s

n1 +n 2 n1 n 2

<− tυ,α

For these data, we have X 1 = .058, s1 = .055, n1 = 23, X 2 = .146, s 2 = .058, and n2 = 23. ( n1 −1) s12 + ( n 2 −1) s 22 ( 22)(.055) 2 + ( 22)(.058) 2 s = = = .0031945 n1 + n 2 − 2 23 + 23 − 2 2

so that s = .0031945 = .0565. Then X1 − X 2 s

n1 + n 2 n1 n 2

=

.058 − .146 .0565

23 + 23 23 × 23

= −5.282

For a 1 percent level test, we have by interpolation from Table for Student’s t distribution with 44(23+23-2) degrees of freedom, t44,.01=2.414. Because -5.282 is much less than -2.414, the null hypothesis is rejected at α=1 percent. The data cast considerable doubt on the hypothesis that the population mean return on assets is at least as large for failed than for nonfailed retail firms. The test just discussed and illustrated is based on the assumption that the two population variances are equal. It is also possible to develop tests that are valid when this assumption does not hold.

Exercise 2.9. Here 36 is chosen as the H0 value even though any number greater than 36 is in H0. The bank’s statistician selects the test statistic as

χ

2 n −1

=

( n −1) s X2

σX2

For a desired significance level of α=.05 and 19 degrees of freedom, Table A5 in Appendix A suggests a critical value of 10.117 (this being a lower-tailed test). Thus the decision rule must be as follows: Fail to reject H0 if

χ 2 ≥ σ 12−.05,19 = 10.117 After taking a sample of 20 customers, the statistician finds the sample single-line waiting times to have a standard deviation of sx=4 minutes per customer. Accordingly, the computed value of the test statistic is

χ n2−1

(n − 1) s X2 4 2 (20 − 1) = = = 8.84 2 36 σX

Because 8.44 is smaller than 10.117, the null hypothesis should be rejected at the 5 percent significance level, which means that the sample result is statistically significant. In other world, the observed divergence from the hypothesized value of σ*X=6 minutes is not likely to be the result of chance factors operating during sampling.

Exercise 2.10.

ˆ = f = 300/2000 = 15%. • p We can consider that p follows a Normal law with mean μ = 0.15 and σˆ * = (0.15 × 0.85) / 2000 = 0.007986 So the required interval of confidence is 0.15 ± (1.96) (0.007986) or P(0.1343 < p < 0.1657) = 0.95 The resulting precision ˆ = (1.96) (0.007986) / (0.15) = 10.4% ε = Δp/ p 2.1) If H0 is true, we have L(fn) = L(X) = B(n;0.25). Here n > 200, so we can approximate this binomial law by a Normal law:

L(fn) ≈ ℵ0 (0.15;

0.15 × 0.85 ) n

Analogously, if H1 is true, we can get

L(fn) ≈ ℵ1 (0.25;

0.25 × 0.75 ) n

The first problem of sampling distribution with the risk accepted α = .01 (tα = 2.3263) gives

2.3263 =

q − 0.15 0.15 × 0.85 n

The second problem of sampling distribution with the risk accepted β = .05 (tβ= -1.6449) gives − 1.6449 =

q − 0.25 0.25 × 0.75 n

From these two equations we get

q − 0.15 2.3263 0.15 × 0.85 = = −1.1662 q − 0.25 − 1.6449 0.25 × 0.75 Therefore q = 0.203837 and after substituting this value of q into the first n = 15.43 → n = 238. equation →

2.2) Now if on the sample of this size we find out f = 22% > q, we should reject H0 and adopt H1. 2.3) With n = 500, if we keep α = .01, we can find from the first equation q = 0.187. Then from the second equation we get tβ= -3.2564, so β < 0.001.