IE 241 Solutions 2

1. Ho: pf = pn

Ha: pf ≠ pn

46/200 = 0.23 fraternity proportion 51/300 = 0.17 nonfraternity proportion 97/500 = 0.19 overall proportion

Because of the large sample size, we can use t.05 = z.05 = 1.96 for α = .05 for this 2-tailed test. Because the observed t < 1.96, we cannot reject the null hypothesis and we conclude that no difference of opinion on this issue has been demonstrated.

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nf

nf

npq

npq

ppt

2. Ho: py – pa = .10 Ha: py – pa > .10

50/200 = .25 youth proportion 60/500 = .12 adult proportion 110/700 = .157 overall proportion

Because of the large sample size, we can use t.

05 = z.05 = 1.65 at α =.05 for this 1-tailed test. Since the observed t < 1.65, we can not reject the null hypothesis.

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10.12.25.

ay

ay

npq

npq

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3. Ho: μA – μB = 0 Ha: μA – μB ≠ 0

sA = 8 sB = 10 nA = 100 nB = 100

For this 2-tailed test at α = .05, t = z = 1.96. Since the observed t > 1.96, we reject Ho and conclude that there is a difference between the teaching methods with 95% confidence.

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B

B

A

A

BA

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ns

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4. Expected ratio: q2 : p2+2pq : r2+2qr : 2pr 0.16 : 0.48 : 0.20 : 0.16

when p = .4, q = .4, r = .2.

Χ2 test of goodness of fit

For 3 df, χ2 = 7.815 at α= .05, so we can reject Ho with 95% confidence and declare that the observed frequencies are not compatible with the pattern.

A B AB O Total

Obs 180 360 132 98 770

Exp

.16(770)=123

.48(770)=370

.2(770)=154

.16(770)=123

770

91.34123

)12398(

154

)154132(

370

)370360(

123

)123180( 22222

5. Ho: Drug and Effect are not related Ha: Drug and Effect are related

χ2 test of independence in contingency table

The expected frequencies are obtained by (ni.)(n.j) / n and are shown in parentheses in the ta

ble.

Helped

Harmed

No effect

Total

Drug 52 (48)

10 (11)

20 (23) 82

Placebo

44 (48)

12 (11)

26 (23) 82

Total 96 22 46 164

5 (continued)

The df here = (2-1)(3-1) = 2 and at α =.05, the critical value of χ2 = 5.99.

Since the observed χ2 < 5.99, we cannot reject Ho.

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)2326(

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)4844(

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6(a). Ho: σ = 8 Ha: σ < 8

= 42 n = 20, s = 5 χ2 test that variance = specified value

The critical value of χ2 for 19 df where α =.05 is 30.1. Since the observed χ2 is much less than 30.1, we cannot reject the null hypothesis. So the data do not support the proposition that the variance is < 8.

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6(b) Ho: μ = 45 Ha: μ ≠ 45 = 42 n = 20, s = 5

The critical value of t for 19 df in this 2-tailed test at α =.05 is 2.093. Since the observed value of t < -2.093, we can reject the null hypothesis that μ = 45.

X

68.2

20

54542

X

Xt

7. Ho: μ1 = μ2

Ha: μ1 ≠ μ2

The t test is

The critical t for (10-1) + (12 -1) = 20 df is 2.086 for this two-tailed test at α = .05. Since the observed t < 2.086, we cannot reject the null hypothesis.

Sample 1 Sample 2

n 10 12

mean 20 24

s 5 6

706.1345.2

4

12

6

10

5

242022

2

22

1

21

21

ns

ns

XXt

7 (continued) Testing the validity of the t test by ensuring that

the two variances are not significantly different.

The critical value of F for (12-1) and (10-1) df at α = .05 is 3.10. Since the observed F < 3.10, the two variances are not significantly different and the t test is valid.

44.15

62

2

22

21 s

sF

8. For a sample of 20 rats, 10 in each group:

Ho: μ raw = μ roasted

Ha: μ raw ≠ μ roasted

The critical t value for (10-1) + (10-1) = 18 df at α = .05 for this two-tailed test is 2.101. Since the observed t < 2.101, we cannot reject the null hypothesis.

Mean Variance

raw 57.9 31.21

roasted 55.8 21.07

918.0286.2

1.2

1007.21

1021.31

8.559.57

2

22

1

21

21

ns

ns

XXt

8 (continued)

Testing the validity of the t test by ensuring that the two variances are not significantly different.

The critical value of F for (10-1) and (10-1) df at α = .05 is 3.18. Since the observed F < 3.18, the two variances are not significantly different and the t test is valid.

48.107.21

21.3122

21 s

sF

9. Is the variability of wood affected by weathering?

The critical value of F = 1.598 for α = .05 for (100-1) and (40-1) df. Since the observed F < 1.598, the data do not justify rejecting the null hypothesis.

Outside

Inside

n 40 100

mean

117 132

s 14.897

16.58924.1

897.14

589.162

2

2

2

out

in

s

sF

10. Ho: μC – μU = 0

Ha: μC – μU ≠ 0

The test statistic is

The critical t value for (20-1)+(20-1) = 38 df at α = .05 for this two-tailed test is -2.024. Since the observed t <-2.024, we can reject the null hypothesis.

Pipes n Mean Variance

Coated 20 44.45 88.47

Uncoated

20 51.4 104.88

235.2109.3

95.6

2088.104

2047.88

4.5145.4422

U

U

C

C

UC

ns

ns

XXt

10 (continued) The 95% confidence interval for the differen

ce between the two means is

So the 95% confidence interval is .66 ≤ μdiff ≤ 13.24

Since this confidence interval does not include 0, it confirms the result of the hypothesis test.

29.695.6)109.3)(024.2(95.6

11. Are gender and color-blindness related?

χ2 test of independence for contingency table The expected frequencies are obtained by (ni.)(n.j)/n, so the expected frequency for cell11 i

s (956)(480)/1000 = 458.88 = 459. The expected frequencies for the other cells are shown in parentheses.

Male Female

Total

Normal 442 514 956

Color-blind

38 6 44

Total 480 520 1000

11 (continued)

The critical value of χ2 for =(2-1)(2-1)=1 df is 3.84. Since the observed χ2 >> 3.84, we can reject the null hypothesis and declare that gender and color-blindness are related.

Male Female Total

Normal 442 (459)

514 (497)

956

Color-blind 38 (21)

6 (23) 44

Total 480 520 1000

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)236(

21

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497

)497514(

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12. 12 pairs of runners matched for ability

Ho: μdiff = 0 Ha: μdiff > 0

The critical t value for (12-1) df for this 1-tailed test at α = .

05 is 1.796. Since the observed t > 1.796, we can reject the null hypothesis and declare the stimulant effective.

Pair 1 2 3 4 5 6 7 8 9 10 11 12

Stimulant

64 62 63 61 61 59 62 64 65 62 63 63

No Stim 64 63 65 62 63 61 62 66 65 64 65 63

Difference

0 1 2 1 2 2 0 2 0 2 2 0

167.1diffX 937.0diffs31.4

12

937.0167.1

t

12 (continued)

The 95% confidence interval for the mean difference is

So the 95% confidence interval is 0.572 < μ < 1.762 Since the interval does not cover 0, it confi

rms the result for the t test.

595.0167.112

937.0)201.2(167.1

13.

Ho: μ1 = μ2

Ha: μ1 < μ2

The critical value of t for 68 df = z for this 1-tailed test at

α = .05 is 1.645. Since the observed t < -1.645, the null hypothesis is rejected and method 1 is declared better (takes less time) than method 2.

Method 1

Method 2

mean 53.7 55.375

variance

15 18

n 30 40

7185.19747.0

675.1

40

18

30

15

7.53375.55

2

22

1

21

21

n

s

n

s

XXt

13 (continued)

The 95% confidence interval for the time difference is

and the interval is

-0.2354 < μ < 3.5854

This interval clearly covers 0 even though the difference is significant because …

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