The following table contains information on matched sample values whose differences are normally distributed.
Number - Sample 1 - Sample 2
1 17 21
2 12 13
3 21 22
4 23 19
5 19 19
6 13 17
7 19 16
8 17 21
a. Construct the 95% confidence interval for the mean difference μD.
SOLUTION:
From given data,
The following table contains information on matched sample values whose differences are normally distributed.
Number | Sample 1 () | Sample 2 () | Difference = - | ( - )2 |
1 | 17 | 21 | -4 | (-4+0.875)2 = 9.765625 |
2 | 12 | 13 | -1 | (-1+0.875)2 = 0.015625 |
3 | 21 | 22 | -1 | (-1+0.875)2 = 0.015625 |
4 | 23 | 19 | 4 | (4+0.875)2 = 23.765625 |
5 | 19 | 19 | 0 | (0+0.875)2 = 0.765625 |
6 | 13 | 17 | -4 | (-4+0.875)2 = 9.765625 |
7 | 19 | 16 | 3 | (3+0.875)2 = 15.015625 |
8 | 17 | 21 | -4 | (-4+0.875)2 = 9.765625 |
= -7 | ( - )2 = 68.875 |
= 8
= -4-1-1+4+0-4+3-4 = -7
The mean of difference = = / = -7 / 8 = -0.875
( - )2
= 9.765625+0.015625+0.015625+23.765625+0.765625+9.765625+15.015625+9.765625
= 68.875
= * ( - )2
= * 68.875
= 9.83928
(a). Construct the 95% confidence interval for the mean difference
Sample standard deviation , S = = = 3.1367626
95% CI for using t-dist
Confidence interval is 95%
95% = 95/100 = 0.95
= 1 - Confidence interval = 1-0.95 = 0.05
/2 = 0.05 / 2 = 0.025
Degree of freedom = df = n-1 = 8-1 = 7
Critical value,
t/2,df = t0.025,7 = 2.3646 ( from t- table , two -tails , df =7 )
Margin of Error = E = t/2,df * ( S / )
= 2.3646 * ( 3.1367626 / )
= 2.3646 *1.109013
= 2.62237
Limits of 95% confidence interval are given by:
Lower limit = - E = -0.875 - 2.62237 = -3.49737 -3.49
Upper limit = + E = -0.875+ 2.62237 = 1.74737 1.74
Therefore, Confidence interval for the mean difference μD
(-3.49 , 1.74 )
-3.49 < D < 1.74
Please thumbs-up / vote up this answer if it was helpful. In case of any problem, please comment below. I will surely help. Down-votes are permanent and not notified to us, so we can't help in that case.
The following table contains information on matched sample values whose differences are normally distributed. Number -...
The following table contains information on matched sample values whose differences are normally distributed. (You may find it useful to reference the appropriate table: z table or t table) Number Sample 1 Sample 2 1 18 22 2 13 11 3 22 23 4 23 20 5 17 21 6 14 16 7 18 18 8 19 20 Construct the 99% confidence interval for the mean difference μD. (Negative values should be indicated by a minus sign. Round intermediate calculations...
The following table contains information on matched sample values whose differences are normally distributed. Use Table 2. Number Sample 1 Sample 2 1 16 20 2 12 13 3 20 22 4 20 22 5 17 20 6 14 16 7 16 18 8 18 21 a. Construct the 99% confidence interval for the mean difference μD. (Negative values should be indicated by a minus sign. Round all intermediate calculations to at...
The following table contains information on matched sample values whose differences are normally distributed. (You may find it useful to reference the appropriate table: z table or t table) Number Sample 1 Sample 2 1 17 20 2 12 12 3 21 22 4 21 20 5 16 21 6 14 16 7 17 18 8 17 20 a. Construct the 90% confidence interval for the mean difference μD. (Negative values should be indicated by a minus sign. Round intermediate...
The following table contains information on matched sample values whose differences are normally distributed. (You may find it useful to reference the appropriate table: z table or t table) NumberSample 1Sample 21162021213320224202251720614167161881821 a. Construct the 99% confidence interval for the mean difference μD. (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.)
The following table contains information on matched sample values whose differences are normally distributed. (You may find it useful to reference the appropriate table: z table or t table) Number Sample 1 16 Sample 2 21 10 12 WN 00 O a. Construct the 90% confidence interval for the mean difference up. (Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answers to 2 decimal places.) Confidence interval is...
A simple random sample of size nis drawn from a population that is normally distributed the sample mean is found to be 113, and the sample standard deviations, is found to be 10 (a) Construct a 95% confidence interval about if the sample size is 22 (b) Construct a 95% confidence interval about the sample on 26 (c) Construct a 90% confidence interval about the sample size is 22 (d) Could we have computed the confidence intervals in parts(a-c) if...
The following information was obtained from matched samples. The daily production rates for a sample of workers before and after a training program are shown below. Worker Before After 1 20 22 2 25 23 3 23 27 4 23 20 5 22 21 6 20 19 7 17 18 8 20 21 9 19 18 Refer to Exhibit 3. Assuming that the population of differences has a normal distribution, what is the degrees of freedom for the t distribution...
A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x̅, is found to be 107 , and the sample standard deviation, s, is found to be 10 .(a) Construct a 98 % confidence interval about μ if the sample size, n, is 22 .(b) Construct a 98 % confidence interval about μ if the sample size, n, is 12 .(c) Construct a 95 % confidence interval about μ if the...
A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x̅, is found to be 109 , and the sample standard deviation, 5 , is found to be 12 .(a) Construct a 96 % confidence interval about μ if the sample size, n, is 23 .(b) Construct a 96 % confidence interval about μ if the sample size, n, is 16 .(c) Construct a 90 % confidence interval about μ if...
The following information was obtained from matched samples taken from two populations. Assume the population of differences is normally distributed. Individual Method 1 Method 2 1 7 5 2 5 9 3 6 8 4 7 7 5 5 6 The 95% confidence interval for the difference between the two population means is