Solutions to STA 4032 Assignment 6

9-10a)  = P( 98.5) + P(> 101.5)

= +

= P(Z 2.25) + P(Z > 2.25) = (P(Z - 2.25)) + (1  P(Z  2.25))

= 0.01222 + 1  0.98778 = 0.01222 + 0.01222 = 0.02444

b)  = P(98.5  101.5 when  = 103)

=

= P(6.75  Z 2.25) = P(Z 2.25)  P(Z 6.75) = 0.01222  0 = 0.01222

c)  = P(98.5  101.5 |  = 105)

=

= P(9.75  Z 5.25) = P(Z 5.25)  P(Z 9.75) = 0  0 = 0

The probability of failing to reject the null hypothesis when it is actually false is smaller in part (c) because the true mean,  = 105, is further from the acceptance region. That is, there is a greater difference between the true mean and the hypothesized mean.

9-14a) P-value = 2(1 -) = 2(1 -) = 2(1 -) = 2(1 - 0.99865) = 0.0027

b) P-value = 2(1 -) = 2(1 -) = 2(1 -) = 2(1 - 0.93319) = 0.13362

c) P-value = 2(1 -) = 2(1 -) = 2(1 -) = 2(1 - 0.99865) = 0.0027

9-36a) P-value = 1- = 1-0.02

b) P-value = 1- = 1-0.97

c) P-value = 1- = 1-0.34

9-42a)

1) The parameter of interest is the true mean water temperature, .

2) H0 :  = 100

3) H1 :  > 100

4)

5) Reject H0 if z0 > z where  = 0.05 and z0.05 = 1.65

6) ,  = 2

7) Because -3.0 < 1.65, fail to reject H0. The mean water temperature is not significantly greater than 100 at  = 0.05.

b) P-value =

c)= (1.65 + 6) = (-4.35) 0

9-44a)

1) The parameter of interest is the true mean melting point, .

2) H0 :  = 155

3) H1 :   155

4)

5) Reject H0 if z0 < z/2 where  = 0.01 and z0.005 = 2.58 or z0 > z/2 where  = 0.01 and z0.005 = 2.58

6)= 154.2,  = 1.5

7) Because –1.69 > -2.58, fail to reject the null hypothesis. There is not sufficient evidence to support the claim the mean melting point differs from 155 F at  = 0.01.

b) P-value = 2P(Z <- 1.69) =2(0.045514) = 0.0910

c)

= (-7.96)- (-13.12) = 0 – 0 = 0

d)

n  2.

9-46 a)

1) The parameter of interest is the true mean tensile strength, .

2) H0 :  = 3500

3) H1 :  3500

4)

5) Reject H0 if z0z/2 where  = 0.01 and z0.005 = 2.58 or z0 > z/2 where  = 0.01 and z0.005 = 2.58

6) ,  = 60

7) Because 2.89 < 2.58, reject the null hypothesis and conclude that the true mean tensile strength differs from 3500 at  = 0.01.

b) Smallest level of significance = P-value = 2[1 (2.89)]= 2[1  .998074] = 0.004. The smallest level of significance at which we reject the null hypothesis is 0.004.

c) = 3470 – 3500 = -30

= (4.312)- (-0.848) = 1 – 0.1982 = 0.8018 so the power = 1-  = 0.20

d)

Therefore, the sample size that should be used equals 47.

e) 99% Confidence Interval

3405.313 3494.687

With 99% confidence, the true mean tensile strength is between 3405.313 psi and 3494.687 psi. We can test the hypotheses that the true mean tensile strength is not equal to 3500 by noting that the value is not within the confidence interval. Hence, we reject the null hypothesis.

9-54a) α = 0.01, n = 20, the critical value = -2.539

b) α = 0.05, n = 12, the critical value = -1.796

c) α = 0.1, n = 15, the critical value = -1.345

9-56a)

b) then

c)

9-60a) degrees of freedom = N – 1 = 16 – 1 = 15

b) SE Mean

c) P-value = 2P(t > 2.8581) = 0.012. We reject the null hypothesis if the P-value < . Thus, we can reject the null hypothesis at significance levels greater than 0.012.

d) If the alternative hypothesis is changed to the one-sided alternative  > 34, the P-value = 0.5(0.12) = 0.006.

e) If the null hypothesis is changed to  = 34.5 versus the alternative hypothesis ( ≠ 34.5) the t statistic is reduced. In particular, and .

Because , we fail to reject the null hypothesis at the 0.05 level of significance.

9-64a)

1) The parameter of interest is the true mean rainfall, .

2) H0 :  = 25

3) H1 :   25

4) t0 =

5) Reject H0 if t0 > t,n-1 where  = 0.01 and t0.01,19 = 2.539 for n = 20

6) = 26.04 s = 4.78 n = 20

t0 =

7) Because 0.97 < 2.539 fail to reject the null hypothesis. There is insufficient evidence to conclude that the true mean rainfall is greater than 25 acre-feet at  = 0.01. The 0.10 < P-value < 0.25.

b) The data on the normal probability plot falls along a line. Therefore, the normality assumption is reasonable.

c) d =

Using the OC curve, Chart VII h) for  = 0.01, d = 0.42, and n = 20, obtain  0.7 and power of 1  0.7 = 0.3.

d) d =

Using the OC curve, Chart VII h) for  = 0.05, d = 0.42, and  0.1 (Power=0.9), n = 75

e) 99% lower confidence bound on the mean diameter

Because the lower limit of the CI is less than 25 there is insufficient evidence to conclude that the true mean rainfall is greater than 25 acre-feet at = 0.01.