SMAM 314 EXAM 3 Name____________________ 1. Mark the following statements true or false. (6 points-­‐2 each) __F___A. A t test for paired differences is appropriate for comparing the means from two independent random samples. __T___B. A null hypothesis that is rejected at α = .01 will be rejected at α = .05 . __T__C. If the p value is .015 the null hypothesis will be rejected at α = .05 but fail to be rejected at α = .01 . 2. Pick out the best choice.(6 points-­‐2 each) ___c__A. A population is known to be normally distributed with known standard deviation 5. There are 25 observations and the sample mean is 20.A 95% confidence interval is a. (18.355,21.645) b. (17.926, 22.064) c.(18.04, 21.96) d. (18.289,21.714) e. (17.424, 22.576). -­‐__d___B. The contents of a sample of 10 16 ounce cans of soda is 16.25. The population standard deviation is unknown. If the test statistic is T =2.262 we would conclude that the average amount of soda in a can is significantly more than 16 ounces at a. α = .05 and α=.01 b. α = .01 but not α=.05 c. any reasonable level of significance d. α = .05 but not α=.01 e. in none of the cases described in a-­‐d. e____C. The p value of a statistical test is .016. The null hypothesis will fail to be rejected for the level of significance a. α = .10 b. α = .05 and α=.01 c. α = .02 d. no alpha (it will be rejected for any level of significance) e. α = .01 and α=.005 3. A random sample of 50 suspension helmets used by motorcycle riders and automobile race car drivers is subjected to an impact test and on 18 of these helmets some damage was observed. A. Find a 95% two sided confidence interval on the true proportion of helmets that would show damage from this test.(8 points)  18
p=
= .36
50
.36 ± 1.96
.36 ± .133
(.227,.493)
(.36)(.64)
50
B. Would you reject H0 p=0.3 in favor of the alternative H1 p≠ 0.3 at α = .05 .Write one sentence explaining why or why not.(4 points) You would fail to reject H0 at α = .05 because 0.3 is contained in the confidence interval. C. How large a sample would be needed if we want to be 95% confident that the error in estimating p is at most 0.02?(8 points) 2
⎛z ⎞ 

N = ⎜ α / 2 ⎟ p(1 − p)
⎝ E ⎠
2
⎛ 1.96 ⎞
N=⎜
(.36)(.64) = 2212.7, N = 2213
⎝ .02 ⎟⎠
or
2
⎛ 1.96 ⎞
N=⎜
(.25) = 2401
⎝ .02 ⎟⎠
4. Cloud seeding has been studied for many decades as a weather modification procedure. The rainfall in acre-­‐feet from 20 clouds that were seeded with silver nitrate is as follows. 18.0
30.7
19.8
27.1
22.3
18.8
31.8
23.4
21.2
27.9
27.1
25.0
24.7
26.9
21.8
29.2
34.8
26.7
31.6
The objective is to perform a test of hypothesis at α = .05 to determine whether a claim that mean rainfall from seeded clouds exceeds 25 acre-­‐feet. Consider the Minitab output. 95% Lower
Variable
N
Mean
C1
20 26.03
StDev
4.78
SE Mean
1.07
Bound
24.18
T
0.97
Answer the following questions.(18 points) A. What is the null and the alternative hypothesis. 31.9
P
0.173
H0 µ = 25 H1 µ > 25 B. What assumptions are needed to perform this test? Normal population Unknown standard deviation Small sample C.What is the region of rejection? T>1.729 C. What is the value of the test statistic and what is the p value? T =.97, pvalue =.173 D. Would you reject or fail to reject H0 at α = .05 ? Fail to rejct H0 E. What conclusion do you come to about the rainfall produced by the clouds? The rainfall produced by the clouds is not significantly greater than 25 acre-­‐feet. 5. Using the Minitab output from problem 4. Variable
C1
N
20
Mean
26.03
StDev
4.78
SE Mean
1.07
Bound
24.18
T
0.97
P
0.173
A. Find a 90% two sided confidence interval on the true standard deviation?(10 points) ⎛ (n − 1)s 2 (n − 1)s 2 ⎞
, 2
⎜
⎟
2
χ1−α / 2 ⎠
⎝ χα / 2
⎛ 19(4.78)2 19(4.78)2 ⎞
⎜ 30.144 , 10.117 ⎟
⎝
⎠
(14.401,42.9099)
(3.794,6.551)
B.Find a 95% upper confidence bound for the true standard deviation.(5 points) UCB=6.551,95% one sided interval (0,6.551) B. At α = .10 what would be your conclusion for a test of hypothesis H0 σ = 8 vs. H1 σ ≠ 8 ? Explain your answer based on the result of part A.(5 points) reject H0 because 8 lies outside the confidence interval. C. At α = .05 what would be your conclusion for a test of hypothesis H0 σ = 20 vs H1 σ < 20 . Explain your answer based on the result of part B.(5 points) Reject H0 because 20 does not lie in the one sided confidence interval. 6. In semiconductor manufacturing wet chemical etching is often used to remove silicon from the backs of wafers prior to metallization. The etch rate is an important characteristic in this process and known to follow a normal distribution. Two different etching solutions have been compared using random samples of 10 wafers for each solution. Summary statistics are computed for the observed etch rate.They are are as follows in mils per minute. Solution 1 Solution 2 x1 = 9.9
x 2 = 10.38
s1 = .421
s 2 = .257
A. Perform a test of hypothesis to determine if there is a statistically significant difference in the etch rates for the two solutions. Use α = .05 . Write the complete report.(20 points) H0 µ1 = µ 2 H1 µ1 ≠ µ 2 Assumptions Independent random samples Normal populations Equal but unknown standard deviations Region of rejection Reject H0 if T>2.101, T<-­‐2.101 Calculation x1 − x 2
(n1 − 1)s12 + (n 2 − 1)s 22
2
T=
,s p =
n1 + n 2 − 2
s p n1 + n1
1
2
9(.421) + 9(.257)2
s =
= .1216,s p = .349 18
9.9 − 10.38 −.48
T=
=
= −3.07
.349 1 + 1 .156
2
2
p
10
10
Reject or fail to reject H0 Reject H0 Conclusion There is evidence of a statistically significant difference in the etch rates. B. Compute s12
F = 2 s2
At α = .10 is there reason to suspect a difference in the variability of the etch rate of the two solutions? Would you come to the same conclusion at α = .05? Formulae (.421)2
F=
= 2.683 (.257)2
F.05 (9,9) = 3.18 You would fail to reject H0 at α = .10 . You would come to the same conclusion at α = .05. 



⎛
p(1 − p) 
p(1 − p) ⎞  x
, p + zα / 2
,⎟ , p = ⎜ p − zα / 2
N
N
N
⎝
⎠
2
⎛z ⎞ 

N = ⎜ α / 2 ⎟ p(1 − p) ⎝ E ⎠
⎛ (n − 1)s 2 (n − 1)s 2 ⎞
, 2
⎜
⎟ 2
χ1−α / 2 ⎠
⎝ χα / 2
x1 − x 2
(n1 − 1)s12 + (n 2 − 1)s 22
2
T=
,s p =
n1 + n 2 − 2
s 1 + 1
p
n1
n2