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The following measurements were recorded for the drying time, in hours, of certa

ID: 3142675 • Letter: T

Question

The following measurements were recorded for the drying time, in hours, of certain brands of latex paint: 3.4 2.5 4.8 2.9 3.6 2.8 3.3 5.6 3.7 2.8 4.4 4.0 5.2 3.0 4.8 a. Assuming that the measurements represent a random sample from a normal population, find a 99% confidence interval for the mean of the drying time. b. Find a 95% production interval for a new observation of drying time. c. Assuming the latex paint comes from two different brands (both normally distributed) and samples are collected as shown in the following table, calculate a 95% confidence interval for for mu_1 - mu_2 if we know that sigma^2_1 = 0.85 and sigma^2_2 = 0.57. Brand A 3.4 2.5 4.8 2.9 3.6 2.8 3.3 5.6 3.7 2.8/Brand B 4.4 4.0 5.2 3.0 4.8 d. With the same data collection and distribution assumptions as in part c., calculate a 95% confidence interval for mu_1 - mu_2 if the population variances are unknown but equal.

Explanation / Answer

Solution

CONFIDENCE INTERVAL FOR MEAN

Let X = drying time of the paint. We are given X ~ N(µ, 2).

100(1 – ) % confidence interval for µ when 2 is unknown is: {Xbar ± (s/n)(t/2)}, where

Xbar = sample mean,

= population standard deviation,

s = sample standard deviation,

n = sample size and

t/2 = upper (/2) % point of t-Distribution with (n - 1) degrees of freedom..

Given n =

15

=

0.01

Xbar =

3.766667

s =

0.982465

t/2 =

2.976843

99% CI for

3.766667

±

0.75514

Lower Bound =

3.011527

Upper Bound =

4.521807

ANSWER

Details of

Excel

computations

Raw Data

i

xi

n

15

1

3.1

Xbar

3.766667

2

2.5

s

0.982465

3

4.8

s^2

0.965238

4

2.9

5

3.6

6

2.8

7

3.3

8

5.6

9

3.7

10

2.8

11

4.4

12

4

13

5.2

14

3

15

4.8

PREDICTION INTERVAL FOR SINGLE OBSERVATION

100(1 – ) % prediction interval for new observation of drying time = {Xbar ± (s.Z/2)}, where Z/2 is the upper /2 percent point of N(0, 1).

So, 95% prediction interval for new observation of drying time = {3.766667 ± (0.982465 x 1.96)}

= 3.767 ± 1.926 ANSWER

Given n =

15

=

0.01

Xbar =

3.766667

s =

0.982465

t/2 =

2.976843

99% CI for

3.766667

±

0.75514

Lower Bound =

3.011527

Upper Bound =

4.521807

ANSWER

Details of

Excel

computations

Raw Data

i

xi

n

15

1

3.1

Xbar

3.766667

2

2.5

s

0.982465

3

4.8

s^2

0.965238

4

2.9

5

3.6

6

2.8

7

3.3

8

5.6

9

3.7

10

2.8

11

4.4

12

4

13

5.2

14

3

15

4.8