Statistical Analysis of Experimental Data: Hypothesis Testing and Confidence Intervals, Assignments of Statistics

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Practice Problems # 09 - Solutions

1. In a test of the effect of dampness on electric connections, 100 electric connections were tested under damp conditions and 150 were tested under dry conditions. Twenty of the damp condition connections failed and only 10 of the dry condition connections failed.

a. Test the hypothesis that the two proportions are equal versus the hypothesis that they are

not equal. Use α = 0.10 and perform the 8 steps for testing hypotheses.

1. The parameters are the unknown population proportions p 1 = proportion of failed connections under damp conditions and p 2 = proportion of failed connections under dry conditions.
2. H 0 : p 1 = p 2
3. H 0 : p 1 ≠ p 2
4. Since the sample sizes are large, we will appeal to the Central Limit Theorem. The test

statistics is 1 2 1 2

1 2

( ˆ^ ˆ ) ( )

ˆ ˆ^1

(1 )

p p p p z p p n n

− − −

=

⎛ ⎞

− ⎜ + ⎟

⎝ ⎠

5. Z has a N(0,1) distribution; so we reject H 0 if and only if zobs > 1.645 or zobs < -1.645.
6. The pooled estimate of failure is given by 1 2 1 2

ˆ 20 10 3

.

Y Y

p n n

+ +

= = = =

+ +

. The

computed test statistic is

(.20 .066667) (0)

.12(1 .12)( )

z obs

− −

= =

− +

7. Since the observed test statistic 3.1782 falls in the rejection region, we reject the hypothesis that the two proportions are equal in favor of the alternative that the two proportions are different. We may have committed a type I error.
8. We have simple random samples from both populations. Since the random samples are large, we can appeal to the Central Limit Theorem and say that Z has a N(0,1) distribution.

b. Find the p-value of this test of hypothesis.

P-value = ( 2 ) ( 1 − Φ ( 3.1782 )) = ( 2 ) ( 0.000741 )= 0.001482. This is certainly smaller than any

reasonable α So H 0 should be rejected. We may have committed a type I error.

c. Find a 90% confidence interval on the difference of the two proportions.

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( )

1 1 2 2 1 2 2 1 1

ˆ 1 ˆ ˆ 1 ˆ .2 .8 0.06667 0.

ˆ ˆ 0.20 0.06667 1.

0.13333 1.645 0.04489 0.059, 0.

p p p p p p z α n n

− −

− ± + = − ± +

= ± =

2. A study in which eight types of white wine reports the tartaric acid concentration (in g/L) measured before and after a cold stabilization process. The results are presented in the following table. Wine Type Before After 1 2.86 2. 2 2.85 2. 3 1.84 1. 4 1.60 1. 5 0.80 0. 6 0.89 0. 7 2.03 1. 8 1.90 1.

Find a 95% confidence interval for the difference of the means for the tartaric concentrations before and after the cold stabilization process. Assume that the measurements of tartaric concentrations follow a normal distribution.

( ) ( ) ( )

( )

2, 1

0.

1.84625 1.6525 2.3646 0.19375 2.3646 0.

0.19375 0.100514 0.093, 0.

w n

s w t n α −

⎛ ⎞

± = − ± ⎜ ⎟= ±

⎝ ⎠

± =

Since the measurements are made on the same wine, this is a paired t type situation; so we use the confidence interval from the paired t. Note: this confidence interval does not contain zero; hence it is plausible that the two means are different from one another.

3. Two sets of data yielded the following statistics: s 12 = 2.3 based on a sample of size 6, and s 22 = 0.6 with a sample of size 10. Find a 99% confidence interval on the ratio of the

standard deviations, σ 1 σ 2.

The confidence interval for the ratio σ 22 σ 12 is given by the following formula:

1 2 1 2

2 2 2 2 2 2 1 2 ; 1, 1 2 2 2 ; 1, (^12) 1 1 1

n n n n

S S

F F

S S

α α

σ σ − − − ≤^ ≤^ − −. We can compute this interval first, take the

reciprocals, and then the square roots. F α (^) 2; n 1 (^) −1, n 2 − 1 = F 0.005;5,9 = 7.47116 and F 1 (^) − α2; n 1 (^) −1, n 2 − 1 = F 0.995;5,9 =0.

The confidence interval on σ 22 σ 12 is

( ) ( ) ( ) ( )

( )

1 2 1 2

2 2 2 2 1 2 ; 1, 1 2 2 ; 1, (^12) 1 1

0.07261 0.6 7.47116 0.

, ,

2.3 2.

0.0189417,1.

n n n n

S S

F F

− α − − (^) S α − − S

⎛ ⎞ ⎛ ⎞

⎜⎜ ⎟⎟ =⎜ ⎟

⎝ ⎠ ⎝^ ⎠

=

To get a confidence interval on σ 12 σ 22 , we take the reciprocals. ( 0.51308, 52.793 ).

To get a confidence interval on σ 1 σ 2 , we take the square roots. ( 0.7163, 7.2659 ).

4. An experiment was performed to determine whether the annealing temperature of ductile iron affects its tensile strength. Five specimens were annealed at each of four temperatures. The tensile strength (in ksi) was measured for each. The partially complete ANOVA table for this problem is as follows:

Source df

Sums of Squares

Mean Squares Temp Error 2. Total 95.

a. Complete the ANOVA table.

b. What is the value of the F statistic for H 0 : μ 1 = μ 2 = μ 3 = μ 4 vs. H a : at least one mean is

different from the others? c. What is the p-value for this test?

a. Complete the ANOVA table.

Analysis of Variance Table Source Sum of Mean Term DF Squares Square temp 3 58.65005 19. Error 16 36.8374 2. Total 19 95.

b. What is the value of the F statistic for H 0 : μ 1 = μ 2 = μ 3 = μ 4 vs. H (^) a: at least one mean is different from the others?

MSTr F MSE

= = =

c. What is the p-value for this test?

p − value = P ⎡⎣^ F 3,16 ≥ 8.49 ⎤⎦=0.