MAT222 Test 2 Fall 2009
1. Out of a random sample of 50 accounts in a certain department store, 40 out of 50 customers had paid their outstanding bills on time for a given month.
(a) Construct a 95% confidence interval of the true proportionof customers who had paid their outstanding bills on time. (10 points)
(b) Test the null hypothesis that p= .6 versus the alternative hypothesis that p >.6 at .05 level of significance. (10 points)
(a).
95% confidence interval for p: = (.8) 1.96(.0565)
=.8 .1108 [.7892, .9108]
(b). H0: p= .6l vs Ha: p >.6
p-value = P(Z > 2.89) < .0019 < 05
Reject H0 at level = .05.
2. A test question is considered good if it differentiates between prepared and unprepared students. The first question on a test was answered correctly by 62 of 80 prepared students and by 26 of 50 unprepared students.
(a) Perform a significance test for the null hypothesis versus where is the population proportion of correct answer for prepared students and is the population proportion of correct answer for unprepared students. What do you conclude? (10 points)
(b) Find a 95% confidence interval for - . (10 points)
(a)
p-value = P(Z > 3.02) = ..0018 < 05
Reject H0 at level = .05; The proportion of correct answer for prepared students is higher than theproportion of correct answer for unprepared students.
(b)
95% confidence interval for p1 - p2: = (.775 - .52) 1.96(.0847)
=.255 .166 [.089, .4221]
3. The following table pertains to a study of the relationship between the speed of promotion and the standard of clothing of bank employees:
Speed of Promotion
Slow Average Fast Total
Poorly dressed 10 18 14 42
Well dressed 16 22 20 58
Total 26 40 34 100
(a)Give the joint distribution of “speed of promotion” and “dressed” for this table. (5 points)
Speed of Promotion
Slow Average Fast
Poorly dressed .10 .18 .14
Well dressed .16 .22 .20
(b)Computer the marginal distribution for the speed of promotion. (5 points)
P(slow) = 26/100 = .26, P(Average) = 40/100 = .40, P(fast)=34/100 = .34
(c)Compute the conditional distribution of the “speed of promotion” for “poorly dressed” and “well dressed”. (5 points)
P(slow|poorly dressed) = 10/42 = .238 P(Average|poorly dressed) = 18/42 = .428 P(fast|poorly dressed)=14/42 = .333.
P(slow|well dressed) = 16/58 = .276 P(Average|well dressed) = 22/58 = .379 P(fast|well dressed)=20/58 = .345.
We wish to use a chi-square test at .05 level of significance to test whether the speed of promotion and the standard of clothing are independence.
(d)State the null and alternative hypotheses for this problem. (5 points)
H0 : Speed of promotion and Clothing are independent/or unrelated.
Ha : Speed of promotion and Clothing are dependent/or related.
.
(e) Test the hypotheses in (d) at 5% level of significance and make a conclusion. (10 points)Expected counts: 10.92, 16.8, 14.28,; 15.08, 23.20, 19.72
degree of freedom = (2 – 1)(3 – 1) = 2
.25 < p-value Fail to reject H0
Speed of promotion and Clothing are independent/or unrelated.
4. (a) Find a 95% confidence interval for the slope in the following setting:
n = 20, , and = 2.2 (5 points)
Use d.f. = 18, t* = 2.093
-5.3 (2.101)(2.2) = -5.3 4.62 = [-9.92, -.68]
(b) Test the null hypothesis that the slope is 0 versus the two-sided alternative. (5 points)
H0: =0 versus Ha: 0, at =.05,
t = -5.3/2.2 = -2.41, p-value = 2P{t19 < -2.41}
d.f. = 18, 2.214 < 2.41 < 2.552 .01 < upper tail probability < .02
Hence .02 < p-value < .04
Reject H0 at level = .05
5. Exercise 2.144 on Page 163 of our textbook gives the modulus of elasticity (MOE) and the modulus of rupture (MOR) for 32 plywood specimens. We regress MOR on the MOE for the 32 specimens. Here is part of the MINITAB output for the regression:
Results for: EX10_021.MTP
Regression Analysis: mor versus moe
The regression equation is
mor = 2653 + 0.00474 moe
Analysis of Variance
Source DF SS MS F P
Regression 1 75573461 75573461 49.30 0.000
Residual Error 30 45987465 1532915.5
Total 31 121560926
(a)Complete the analysis of variance table by filling in the “Residual Error” row. (5 points)
(b)What are the values of the regression standard errors s and the squared correlation r2? (5 points)
S = square root of MSE =1239.1; r2 = SSM/SST = 75573461/121560926 = .62
(c)The standard deviation of the moe is 325293. Find the standard error for the least-squares slope b1. (5 points)
.000695
(d) Give a 95% confidence interval for the slope 1 of the regression line. (5 points)
d.f =30, t* = 2.042
95% confidence interval for the slope 1: