Confidence Intervals for Population Means and Differences, Lecture notes of Mathematics

Download Confidence Intervals for Population Means and Differences and more Lecture notes Mathematics in PDF only on Docsity!

Chapter 4

9.2 An electrical firm manufactures light bulbs that have a length of life that is

approximately normally distributed with a standard deviation of 40 hours. If

a sample of 30 bulbs has an average life of 780 hours, find a 96% confidence interval for

the population mean of all bulbs produced by this firm.

Population normal and 𝝈 = 𝟒𝟎 ”known ” , 𝑛 = 30 , 𝑋

̅

= 780

𝟗𝟔% 𝑪. 𝑰 𝒇𝒐𝒓 𝝁 is: 𝑿

̅

± 𝒁

𝟏−

𝜶

𝟐

𝝈

√

𝒏

𝟏)𝜶 = 𝟏 − 𝟎. 𝟗𝟔 = 𝟎. 𝟎𝟒 , 𝟐) 𝟏 −

𝜶

𝟐

= 𝟏 −

𝟎.𝟎𝟒

𝟐

= 𝟎. 𝟗𝟖 , 𝟑)𝒁

𝟏−

𝜶

𝟐

= 𝒁

𝟎.𝟗𝟖

= 𝟐. 𝟎𝟓

𝟒) 𝟕𝟖𝟎 ± 𝟐. 𝟎𝟓

𝟒𝟎

√𝟑𝟎

>> 𝟕𝟖𝟎 ± 𝟏𝟒. 𝟗𝟕𝟏

( 𝟕𝟖𝟎 − 𝟏𝟒. 𝟗𝟕𝟏, 𝟕𝟖𝟎 + 𝟏𝟒. 𝟗𝟕𝟏) = (𝟕𝟔𝟓. 𝟎𝟑 , 𝟕𝟗𝟒. 𝟗𝟕𝟏)

9.6 How large a sample is needed in Exercise 9.2 if we wish to be 96% confident that our

sample mean will be within 10 hours of the true mean?

𝒏 = (

𝒁

𝟏−

𝜶

𝟐

. 𝝈

𝒆

)

𝟐

= (

𝟐. 𝟎𝟓

(

𝟒𝟎

)

𝟏𝟎

)

𝟐

= 𝟔𝟕. 𝟐𝟒 ≈ 𝟔𝟖

“we always rounded the number up”

9.4 The heights of a random sample of 50 college students showed a mean of 174.

centimeters and a standard deviation of 6.9 centimeters.

(a) Construct a 98% confidence interval for the mean height of all college students.

(b) What can we assert with 98% confidence about the possible size of our error if we

estimate the mean height of all college students to be 174.5 centimeters?

𝑛 = 50 , 𝑋

̅

= 174. 5 , 𝑆 = 6. 9

(

𝝈 𝒖𝒏𝒌𝒏𝒐𝒘𝒏

)

a) 𝟗𝟖% 𝑪. 𝑰 𝒇𝒐𝒓 𝝁 is: 𝑿

̅

± 𝒁

𝟏−

𝜶

𝟐

𝑺

√

𝒏

𝟏)𝜶 = 𝟏 − 𝟎. 𝟗𝟖 = 𝟎. 𝟎𝟐 , 𝟐) 𝟏 −

𝜶

𝟐

= 𝟏 −

𝟎. 𝟎𝟐

𝟐

= 𝟎. 𝟗𝟗 , 𝟑)𝒁

𝟏−

𝜶

𝟐

= 𝒁

𝟎.𝟗𝟗

= 𝟐. 𝟑𝟑

𝟒) 𝟏𝟕𝟒. 𝟓 ± 𝟐. 𝟑𝟑

𝟔.𝟗

√𝟓𝟎

>> 𝟏𝟕𝟒. 𝟓 ± 𝟐. 𝟐𝟕𝟑𝟔

𝟗𝟖% 𝑪. 𝑰 𝒇𝒐𝒓 𝝁 ∈ (𝟏𝟕𝟐. 𝟐𝟑 , 𝟏𝟕𝟔. 𝟕𝟕)

b) The error will not exceed 𝒁

𝟏−

𝜶

𝟐

𝑺

√

𝒏

= 𝟐. 𝟐𝟕𝟑𝟔

H.W 9.5 A random sample of 100 automobile owners in the state of Virginia shows that

an automobile is driven on average 23,500 kilometers per year with a standard deviation

of 3900 kilometers. Assume the distribution of measurements to be approximately

normal.

(a) Construct a 99% confidence interval for the average number of kilometers an

automobile is driven annually in Virginia.

(b) What can we assert with 99% confidence about the possible size of our error if

we estimate the average number of kilometers driven by car owners in Virginia to

be 23,500 kilometers per year?

𝑛 = 100 , 𝑋

̅

= 23500 , 𝑆 = 3900

(

𝝈 𝒖𝒏𝒌𝒏𝒐𝒘𝒏

)

a) 𝟗𝟗% 𝑪. 𝑰 𝒇𝒐𝒓 𝝁 is: 𝑿

̅

± 𝒁

𝟏−

𝜶

𝟐

𝑺

√𝒏

𝟏)𝜶 = 𝟏 − 𝟎. 𝟗𝟗 = 𝟎. 𝟎𝟏 , 𝟐) 𝟏 −

𝜶

𝟐

= 𝟏 −

𝟎. 𝟎𝟏

𝟐

= 𝟎. 𝟗𝟗𝟓 , 𝟑)𝒁

𝟏−

𝜶

𝟐

= 𝒁

𝟎.𝟗𝟗𝟓

= 𝟐. 𝟓𝟕𝟓

𝟒) 𝟐𝟑𝟓𝟎𝟎 ± 𝟐. 𝟓𝟕𝟓

𝟑𝟗𝟎𝟎

√

𝟏𝟎𝟎

>> 𝟐𝟑𝟓𝟎𝟎 ± 𝟏𝟎𝟎𝟒. 𝟐𝟓

𝟗𝟗% 𝑪. 𝑰 𝒇𝒐𝒓 𝝁 ∈ (𝟐𝟐𝟒𝟗𝟓. 𝟕𝟓, 𝟐𝟒𝟓𝟎𝟒. 𝟐𝟓)

b) The error will not exceed 𝒁

𝟏−

𝜶

𝟐

𝑺

√𝒏

= 𝟏𝟎𝟎𝟒. 𝟐𝟓

Q. A group of 10 college students were asked to report the number of hours that they spent

doing their homework during the previous weekend and the following results were

obtained: 7.25, 8.5, 5.0, 6.75, 8.0, 5.25, 10.5, 8.5, 6.75, 9.

It is assumed that this sample is a random sample from a normal distribution with unknown

variance 

2

. Let  be the mean of the number of hours that the college student spend doing

his/her homework during the weekend.

(a) Find the sample mean and the sample variance.

𝑋

̅

= 7. 575 , 𝑆

2

= ( 1. 724 )

2

(

𝜎

2

𝑢𝑛𝑘𝑛𝑜𝑤𝑛

)

(b) Find a point estimate for 

𝑋

̅

= 7. 575

(c) Construct a 80% confidence interval for .

𝑋

̅

= 7. 575 , 𝑆 = 1. 724 (𝜎 𝑢𝑛𝑘𝑛𝑜𝑤𝑛) , 𝑑𝑓 = 𝑛 − 1 = 9

80% 𝐶. 𝐼 for is: 𝑋

̅

± 𝑡

𝛼

2

𝑆

√

𝑛

1 ) 𝛼 = 1 − 0. 80 = 0. 20 2 )

𝛼

2

=

1. 20

2

= 0. 1 3 ) 𝑡

𝛼

2

= 𝑡

1. 1

= 1. 383

4 ) 7. 575 ± 1. 383

1. 724

√

10

7. 575 ± 0. 754 (error=e=0.754)

( 7. 575 − 0. 754 , 7. 575 + 0. 754 ) = (6.821, 8.329)

9.35 A random sample of size n 1 = 25 , taken from a normal population with a standard

deviation σ 1

= 5 , has a mean 𝑋

̅

1

= 𝟖𝟎. A second random sample of size n 2

= 36 , taken

from a different normal population with a standard deviation σ 2

= 3 , has a mean 𝑋

̅

2

= 𝟕𝟓.

Find a 94% confidence interval for μ 1

− μ 2

.

𝟗𝟒% 𝑪. 𝑰 𝒇𝒐𝒓 𝝁

𝟏

− 𝝁

𝟐

𝒊𝒔 : (𝑿

̅

𝟏

− 𝑿

̅

𝟐

) ± 𝒁

𝟏−

𝜶

𝟐

√

𝝈

𝟏

𝟐

𝒏

𝟏

+

𝝈

𝟐

𝟐

𝒏

𝟐

𝟏) 𝜶 = 𝟏 − 𝟎. 𝟗𝟒 = 𝟎. 𝟎𝟒 𝟐) 𝟏 −

𝜶

𝟐

= 𝟏 −

𝟎.𝟎𝟒

𝟐

= 𝟎. 𝟗𝟕 𝟑) 𝒁

𝟏−

𝜶

𝟐

= 𝒁

𝟎.𝟗𝟕

= 𝟏. 𝟖𝟖

𝟒) (𝟖𝟎 − 𝟕𝟓) ± 𝟏. 𝟖𝟖

√

𝟓

𝟐

𝟐𝟓

+

𝟑

𝟐

𝟑𝟔

≫ 5 ± 𝟐. 𝟏𝟎𝟏𝟗 (error=e=2.1019)

𝝁

𝟏

− 𝝁

𝟐

∈ (2.8981, 7.1019)

9.38 Two catalysts(محفز) in a batch chemical process, are being compared for their effect

on the output of the process reaction. A sample of 12 batches was prepared using catalyst

1 , and a sample of 10 batches was prepared using catalyst 2. The 12 batches for which

catalyst 1 was used in the reaction gave an average yield of 85 with a sample standard

deviation of 4 , and the 10 batches for which catalyst 2 was used gave an average yield of

81 and a sample standard deviation of 5.

Find a 90% confidence interval for the difference between the population means,

assuming that the populations are approximately normally distributed with equal

variances.

𝒏

𝟏

= 𝟏𝟐, 𝑿

̅

𝟏

= 𝟖𝟓, 𝒔

𝟏

= 𝟒

𝒏

𝟐

= 𝟏𝟎 , 𝑿

̅

𝟐

= 𝟖𝟏, 𝒔

𝟐

= 𝟓

𝝈

𝟏

𝟐

𝒂𝒏𝒅 𝝈

𝟐

𝟐

𝒖𝒏𝒌𝒏𝒐𝒘𝒏 but equal

𝟗𝟎% 𝑪. 𝑰 𝒇𝒐𝒓 𝝁

𝟏

− 𝝁

𝟐

𝒊𝒔 :

(

𝑿

̅

𝟏

− 𝑿

̅

𝟐

)

± 𝒕

𝜶

𝟐

,𝒏 𝟏

+𝒏 𝟐

−𝟐

𝒔

𝒑

√

𝟏

𝒏 𝟏

+

𝟏

𝒏 𝟐

𝟏) 𝜶 = 𝟏 − 𝟎. 𝟗𝟎 = 𝟎. 𝟏 𝟐)

𝜶

𝟐

=

𝟎. 𝟏

𝟐

= 𝟎. 𝟎𝟓 𝟑) 𝒕

𝜶

𝟐

= 𝒕

𝟎.𝟎𝟓

= 𝟏. 𝟕𝟐𝟓

𝑺

𝒑

=

√

(

𝒏

𝟏

− 𝟏

)

𝑺

𝟏

𝟐

+

(

𝒏

𝟐

− 𝟏

)

𝑺

𝟐

𝟐

𝒏

𝟏

+ 𝒏

𝟐

− 𝟐

=

√

𝟏𝟏

(

𝟏𝟔

)

+ 𝟗(𝟐𝟓)

𝟐𝟎

= 𝟒. 𝟒𝟕𝟕𝟕

𝟒)

(

𝟖𝟓 − 𝟖𝟏

)

±

(

𝟏. 𝟕𝟐𝟓

)(

𝟒. 𝟒𝟕𝟕𝟕

)√

𝟏

𝟏𝟐

+

𝟏

𝟏𝟎

≫ 4 ± 𝟑. 𝟑𝟎𝟕𝟐

(𝒆𝒓𝒓𝒐𝒓 = 𝒆 = 𝟑. 𝟑𝟎𝟕𝟐 )

𝝁

𝟏

− 𝝁

𝟐

∈ (0.693, 7. 307 )

H.W 9.41 The following data represent the length of time, in days, to recovery for

patients randomly treated with one of two medications to clear up severe bladder

infections:

Medication 1 Medication 2

𝑛

1

= 14 𝑛

2

= 16

𝑥̅

1

= 17 𝑥̅

2

= 19

𝑠

1

2

= 1. 5 𝑠

2

2

= 1. 8

Find a 99% confidence interval for the difference 𝝁 𝟐

− 𝝁

𝟏

𝝈

𝟏

𝟐

𝒂𝒏𝒅 𝝈

𝟐

𝟐

𝒖𝒏𝒌𝒏𝒐𝒘𝒏 but equal

𝟗𝟎% 𝑪. 𝑰 𝒇𝒐𝒓 𝝁

𝟐

− 𝝁

𝟏

𝒊𝒔 :

(

𝑿

̅

𝟐

− 𝑿

̅

𝟏

)

± 𝒕

𝜶

𝟐

,𝒏 𝟏

+𝒏 𝟐

−𝟐

𝒔

𝒑

√

𝟏

𝒏

𝟏

+

𝟏

𝒏

𝟐

𝟏) 𝜶 = 𝟏 − 𝟎. 𝟗𝟗 = 𝟎. 𝟎𝟏 𝟐)

𝜶

𝟐

=

𝟎. 𝟎𝟏

𝟐

= 𝟎. 𝟎𝟎𝟓 𝟑) 𝒕

𝜶

𝟐

= 𝒕

𝟎.𝟎𝟎𝟓

= 𝟐. 𝟕𝟔𝟑

𝑺

𝒑

= √

(𝒏

𝟏

− 𝟏)𝑺

𝟏

𝟐

+ (𝒏

𝟐

− 𝟏)𝑺

𝟐

𝟐

𝒏

𝟏

+ 𝒏

𝟐

− 𝟐

=

√

𝟏𝟑 (𝟏. 𝟓

𝟐

) + 𝟏𝟓(𝟏. 𝟖

𝟐

)

𝟐𝟖

= 𝟏. 𝟑𝟑𝟑𝟔

𝟒)

(

𝟏𝟗 − 𝟏𝟕

)

±

(

𝟐. 𝟕𝟔𝟑

)(

𝟏. 𝟑𝟑𝟑𝟔

)√

𝟏

𝟏𝟒

+

𝟏

𝟏𝟔

≫ 2 ± 𝟏. 𝟑𝟒𝟖

(𝒆𝒓𝒓𝒐𝒓 = 𝒆 = 𝟏. 𝟑𝟒𝟖 )

𝝁

𝟐

− 𝝁

𝟏

∈ (0.6 5 , 3.35)

9.44 A taxi company is trying to decide whether to purchase brand "A" or brand "B" tires

for its fleet of taxis (اسطول من سيارات التكسي). The experiment is conducted using 12 of each

brand and the tires are run until they wear out.

I) Compute a 9 9 % confidence interval for μ 1

− μ 2

, assuming the populations to

be approximately normally.

II) Find a 99% confidence interval for μ 1

− μ 2

if tires of the two brands are

assigned at random to the left and right rear wheels of 8 taxis and the

following distances, in kilometers, are recorded:

Taxi Brand A Brand B

1 34,400 36,

2 45,500 46,

3 36,700 37,

4 32,000 31,

5 48,400 47,

6 32,800 36,

7 38,100 38,

8 30,100 31,50 0

Assume that the differences of the distances are approximately normally distributed.

𝝈

𝟏

𝟐

𝒂𝒏𝒅 𝝈

𝟐

𝟐

𝒖𝒏𝒌𝒏𝒐𝒘𝒏, 𝝈

𝟏

𝟐

≠ 𝝈

𝟐

𝟐

.

I)

From the table, we calculate:

𝑿

̅

𝟏

= 𝟑𝟕, 𝟐𝟓𝟎, 𝒔

𝟏

= 𝟔𝟓𝟒𝟔. 𝟕𝟓𝟓

𝑿

̅

𝟐

= 𝟑𝟖, 𝟑𝟔𝟐. 𝟓, 𝒔

𝟐

= 𝟔𝟏𝟖𝟏. 𝟎𝟔𝟑

𝟗𝟗% 𝑪. 𝑰 𝒇𝒐𝒓 𝝁

𝟏

− 𝝁

𝟐

𝒊𝒔 :

(

𝑿

̅

𝟏

− 𝑿

̅

𝟐

)

± 𝒕

𝜶

𝟐

,𝒏

𝟏

+𝒏

𝟐

−𝟐

√

𝒔

𝟏

𝟐

𝒏

𝟏

+

𝒔

𝟐

𝟐

𝒏

𝟐

𝟏)𝜶 = 𝟏 − 𝟎. 𝟗𝟗 = 𝟎. 𝟎𝟏 , 𝟐)

𝜶

𝟐

=

𝟎. 𝟎𝟏

𝟐

= 𝟎. 𝟎𝟎𝟓 , 𝟑)𝒕𝜶

𝟐

= 𝒕

𝟎.𝟎𝟎𝟓,𝟏𝟒

= 𝟐. 𝟗𝟕𝟕

𝟒) (𝟑𝟕𝟐𝟓𝟎 − 𝟑𝟖𝟑𝟔𝟐. 𝟓) ± (𝟐. 𝟗𝟕𝟕)

√

𝟔𝟓𝟒𝟔. 𝟕𝟓𝟓

𝟐

𝟖

+

𝟔𝟏𝟖𝟏. 𝟎𝟔𝟑

𝟐

𝟖

≫ −𝟏𝟏𝟏𝟐. 𝟓 ± 𝟗𝟒𝟕𝟔. 𝟓𝟖𝟕

(𝒆𝒓𝒓𝒐𝒓 = 𝒆 = 𝟐𝟎𝟓𝟒𝟔𝟗 )

𝝁

𝟐

− 𝝁

𝟏

∈ ( - 10589.0873 , 8364.0873 )

II)

𝟗𝟗% 𝑪. 𝑰 𝒇𝒐𝒓 𝝁

𝒅

𝒊𝒔 ∶ [ 𝒅

̅

± 𝒕∝

𝟐

,𝒏−𝟏

𝑺

𝒅

√

𝒏

]

𝒅

̅

=

∑

𝒅

𝒊

𝒏

= , 𝑺

𝒅

= 𝟏𝟒𝟓𝟒. 𝟒𝟖𝟖

𝟏)𝜶 = 𝟏 − 𝟎. 𝟗𝟗 = 𝟎. 𝟎𝟏 , 𝟐)

𝜶

𝟐

=

𝟎. 𝟎𝟓

𝟐

= 𝟎. 𝟎𝟎𝟓 , 𝟑)𝒕𝜶

𝟐

,𝒏−𝟏

= 𝒕

𝟎.𝟎𝟎𝟓,𝟕

= 𝟑. 𝟒𝟗𝟗

𝟒) −𝟏𝟏𝟏𝟐. 𝟓 ±

(

𝟑. 𝟒𝟗𝟗

)

(

𝟏𝟒𝟓𝟒.𝟒𝟖𝟖

√

𝟖

) ≫ −𝟏𝟏𝟏𝟐. 𝟓 ± 𝟏𝟕𝟗𝟗. 𝟑

𝝁

𝒅

∈ [ −𝟐𝟗𝟏𝟏. 𝟖 , 𝟔𝟖𝟔. 𝟖 ]

Values of Z

𝒁

𝟎.𝟗𝟎

𝒁

𝟎.𝟗𝟓

𝒁

𝟎.𝟗𝟕

𝒁

𝟎.𝟗𝟕𝟓

𝒁

𝟎.𝟗𝟖

𝒁

𝟎.𝟗𝟗

𝒁

𝟎.𝟗𝟗𝟓

H.W

Q2. Suppose that we are interested in making some statistical inferences about the mean,

, of a normal population with standard deviation =2.0. Suppose that a random sample of

size n =49 from this population gave a sample mean =4.5. X

Q1. A survey of 500 students from a college of science shows that 275 students own

(1) The distribution of X is.

(2) A good point estimate of  is = 𝑿

̅

= 𝟒. 𝟓

(3) The standard error of X is = 0.

(4) A 95% confidence interval for  is (𝟑. 𝟗𝟒 , 𝟓. 𝟎𝟔)

(5) If the upper confidence limit of a confidence interval is 5.2 , then the lower confidence limit is

(6) The confidence level of the confidence interval (3.88, 5.12) is

𝑿

̅

+ 𝒁

𝟏−

𝜶

𝟐

𝝈

√

𝒏

= 𝟓. 𝟏𝟐 >> 𝟒. 𝟓 + 𝒁

𝟏−

𝜶

𝟐

𝟐

√𝟒𝟗

= 𝟓. 𝟏𝟐

𝒁

𝟏−

𝜶

𝟐

=

(

𝟓. 𝟏𝟐 − 𝟒. 𝟓

)

√𝟒𝟗

𝟐

= 𝟐. 𝟏𝟕

𝑷

(

𝒁 < 𝟐. 𝟏𝟕

)

= 𝟏 −

𝜶

𝟐

= 𝟎. 𝟗𝟖𝟓 ≫ 𝜶 = 𝟐

(

𝟎. 𝟎𝟏𝟓

)

= 𝟎. 𝟎𝟑

Then, the confidence level = 𝟏 − 𝜶 = 𝟎. 𝟗𝟕 ≫ 𝟎. 𝟗𝟕

(

𝟏𝟎𝟎

)

= 𝟗𝟕%

Note: we will get the same result if use 𝑿

̅

− 𝒁

𝟏−

𝜶

𝟐

𝝈

√

𝒏

= 𝟑. 𝟖𝟖

(7) If we use X to estimate , then we are 95% confident that our estimation error will not exceed. 𝒆 =

𝟎. 𝟓𝟔

(8) If we want to be 95% confident that the estimation error will not exceed e=0.1 when we use X to

estimate , then the sample size n must be equal to

𝒏 = 𝟏𝟓𝟑𝟕

computers. In another independent survey of 400 students from a college of engineering

shows that 240 students own computers.

(a) a 99% confidence interval for the true proportion of college of science's student who

own computers is ( 𝟎. 𝟒𝟗𝟐𝟕, 𝟎. 𝟔𝟎𝟕𝟑)

(b) a 95% confidence interval for the difference between the proportions of students

owning computers in the two colleges is

( −𝟎. 𝟏𝟏𝟒𝟖, 𝟎. 𝟎𝟏𝟒𝟖)