Practice Problems 8 Solutions - Applied Statistics for Engineers and Science | STAT 541, Assignments of Statistics

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Practice Problems # 08 - Solutions

1. Gravel pieces are classified as small, medium, or large. A vendor claims that at least 10% of the gravel pieces from her plant are large. In a random sample of 1600 pieces, 150 pieces were classified as large. Is this enough evidence to reject the claim?

This is a very large sample size, and is a test of hypothesis involving a proportion. Therefore, we are quite comfortable in using the Central Limit Theorem. Thus, this is a z-test. H 0 : p = 0. H (^) a : p < 0. Rejection Region: zobs < -zα.

( ) ( )

0 0 0

ˆ 150 1600 0.10 0.09375 0.

0.

1 0.10 1 0.10 0.

p p z p p n

− − −

= = = = −

− −

The p-value for this is P Z [ ≤ −0.833 ]= 0.202; this p-value is certainly larger than any

reasonably chosen significance level. Therefore, we fail to reject H 0 ; we may have committed a type II error. No, this is not enough evidence to reject her claim.

2. A new gun-like apparatus has been created by a medical engineer to replace the needle in administering vaccines. The apparatus, which is connected to a large supply of the vaccine, can be set to inject different amounts of the serum, but the variance in the amount of serum injected to a given person must not be greater than .06 to ensure proper inoculation. A random sample of twenty-five injections resulted in a sample variance of 0.135 (i.e. s^2 = 0.135).

a. Find a 90% confidence interval for the true variance σ^2.

b. Does is matter whether or not the sample (the 25 observations used in this experiment) is a random sample from a normally distributed population? Explain. c. Do the data provide evidence to indicate the gun is not working properly? To answer this question, interpret the confidence interval computed in part a. as an

interval of plausible values for the unknown parameter σ^2.

a. n = 25 (n-1) = 24 degrees of freedom for the Chi-Square s 2 = 0.135 α = 0.10 α/2 = 0.05 1-α/2 =.

χ 1 − α 2, n − 1 = χ.95,24 =13.

χ α 2, n − 1 = χ.05,24 =36.

2 2

2, 1 1 2, 1

( 1)( ) ( 1)( ) (24)(0.135) (24)(0.135)

, ,

n n 36.415^ 13.

n s n s

χ α − χ− α −

⎛ − − ⎞ ⎛ ⎞

⎜⎜ ⎟⎟ =⎜ ⎟

⎝ ⎠ ⎝^ ⎠

= (0.08897 , 0.23397)

b. Yes, it is necessary to assume that the observations come form a normally

distributed population in order for the statistic

2 2

( n 1) S

σ

−

to have a Chi-Squared

distribution with n-1 d.f. Individuals in the past have pointed out that the sample cannot be normally distributed, and that is true. The histogram of the observed values will never be exactly normally distributed. What we must assume here is that the

random sample was taken from a normally distributed population. In statistical parlance, we use the phrase "the sample must be normally distributed", and what we mean is that the random variables X 1 , X 2 , … , Xn are each normally distributed, which is equivalent to saying that the random sample comes from a normally distributed population.

c. Yes, the upper specification limit of 0.06 for the variance of the dispensed dose is outside (below) the 90% confidence interval; hence it is not a plausible value for the population variance. We therefore infer that the gun is not working within specification limits.

Some authors use the phrase “is there sufficient evidence in the data to say the gun is not working properly?” What is of importance in this context is what we mean by "sufficient evidence to indicate". In statistical inference, when the value of question is not a plausible value (outside the confidence interval), we interpret this to mean that there is indeed sufficient evidence in the data to indicate that this value of interest is not plausible. That is, being outside the confidence interval is considered sufficient evidence that the value is not plausible. Therefore, it is reasonable to conclude that the gun is producing variation in the amount of injected vaccine that is too large; sometimes it may be too much and others too little. We conclude the gun is not working properly.

One must ALWAYS remember that sufficient or significant statistical evidence does not imply practical significance.

3. Measurements of ammonium concentrations (in mg/L) at a large number of wells were made in the stat of Iowa. These included 349 alluvial wells and 143 quaternary wells. The concentrations in the alluvial wells gave a sample average of 0.27, and the assumed population standard deviation

for alluvial wells is σ 1 = 0.40. Those at the quaternary wells have a sample average of 1.62 with σ 2

= 1.70. Find a 95% confidence interval for the difference in mean concentration between alluvial and quaternary wells.

( ) ( ) ( )

( ) ( ) ( ) ( )

( )

2 2 2 2 1 2 2 1 2

0.40 1.

0.27 1.62 1.9600 1.35 1.96 0.

1.35 0.28178 1.632, 1.

x y z α n n

σ σ

− ± + = − ± + = − ±

= − ± = − −

4. The carbon content (in parts per million – ppm) was measured five times for each of two different silicon wafers. The measurements were as follows: Wafer A: 1.10 1.15 1.16 1.10 1. Wafer B: 1.20 1.18 1.16 1.18 1.

Test the hypothesis that the two means are equal using the Welch’s T’ and α = 0.01.

1. The parameters are μ 1 = the mean carbon content of wafer type A, and μ 2 = the mean carbon content of wafer type B.
2. H 0 : μ 1 = μ 2

3. H (^) a : μ 1 ≠ μ 2
4. Test statistic is Welch’s approximate T’; 1 2 2 2 1 2 1 2

( X Y ) ( )

S S

n n

− − μ −μ

⎛ ⎞

⎜ + ⎟

⎝ ⎠

5. We reject H 0 in favor of Ha if and only if tobs ' ≥ t α 2,ν or tobs ' ≤ − t α 2,ν 2 1 2 2 2 1 2 1 2

( )

V V

V V

n n

ν

+

=

+

− −

where

2 1 1 1

S

V

n

= and

2 2 2 2

S

V

n

=

2 S 1 (^) = 0.00080and V 1 =. 2 S 2 (^) = 0.00038and V 2 =. ( ) ( ) ( )

2 2 1 2 2 2 2 2 1 2 1 2

( ) 0.00016^ 0.000076 0.

0.00016 0.000076 0.0000000064^.

V V

V V

n n

ν

+ +

= = = =

+

+ +

− −

t 0.005,7.1 (^) =3.

6. '^2 1

1 2 1 2

1.13 1.

( ) ( ) (1.13 1.174) (0)

.0008.

( ) (^ )

X Y

X Y

T

S S

n n

μ μ

= =

− − − − −

= = = −

+ +

7. Since -2.8642 > -3.48276, we fail to reject the null hypothesis. The two means may indeed be equal. We may have committed a type II error.
8. Both samples must be random samples from the two populations, and the samples must be independent of each other. Likewise, since n = 5 is a small sample size, we must assume that the two populations are normally distributed.