poisonroxs
New member
- Joined
- Apr 12, 2008
- Messages
- 12
Hoping someone could let me know if this is correct. I feel confident in a and b however I am not as confident about c and d. Any input is appreciated, thank you in advance 
)
A sample of 20 pages was taken without replacement from the 1,591-page phone directory. On each page, the mean area devoted to display ads was measured. The data, in square millimeters is shown below:
0 260 356 403 536 0 268 369 428 536
268 396 469 536 162 338 403 536 536 130
A) Construct a 95% confidence interval for the true mean.
B) Why might normality be an issue here?
C) What sample size would be needed to obtain an error of ± 10 square millimeters with 99% confidence?
D) If this is not a reasonable requirement, suggest one that is.
A)
Number of cases =20
Mean or Xbar = 346.5
Squared deviations=
(0-346.5)²= (-346.5)²= 120062.25
(260-346.5)²= (-86.5)² = 7482.25
(356-346.5)² = (9.5)² = 90.25
(403-346.5)² = (56.5)² = 3192.25
(536-346.5)² = (189.5)² = 35910.25
(0-346.5)² = (-346.5)² =120062.25
(268-346.5)²= (-78.5)² =6162.25
(369-346.5)² = (22.5)² =506.25
(428-346.5)² = (81.5)² = 6642.25
(536-346.5)² = (189.5)² =35910.25
(268-346.5)² = (-78.5)² = 6162.25
(396-346.5)² = (49.5)² =2450.25
(469-346.5)² = (122.5)²=15006.25
(536-346.5)² = (189.5)² =35910.25
(162-346.5)²= (-184.5)²=34040.25
(338-346.5)² = (-8.5)²= 72.25
(403-346.5)²= (56.5)²=3192.25
(536-346.5)²= (189.5)²=35910.25
(536-346.5)²= (189.5)² = 35910.25
(130-346.5)²= (-216.5)² =46872.25
Variance = 551547/19
Variance= 29028.7895
Standard deviation =sqrt (variance)
=sqrt (551547/19)
S= 170.3784
Xbar (sample mean) = 346.5
N=20
Se=(standard error)= sd/sqrt(n)= 38.0978
t= 2.093 (refers to the t-distribution critical value with 19 degrees of freedom)
xbar-t se=346.5 –(2.093)(38.0978)=
346.5-79.739=266.76
Xbar +t se= 346.5 + (2.093)(38.0978)=
346.5 + 79.739=426.24
266.76 <µ<426.24
266.76<346.5<426.24
B)
There is a small sample and the data is not continuous since there are two zeros. The normality can be tested by entering the data and frequency to see if the data is normal. Zero has a frequency of 2.536
C)
N=(2.57 sd/10)²=1917 samples
2.57 corresponds to 99% confidence assuming normality.
D)
99% confidence is okay assuming normality. If the distribution is not normal, reducing it to 90% confidence would reduce the required sample size.
)A sample of 20 pages was taken without replacement from the 1,591-page phone directory. On each page, the mean area devoted to display ads was measured. The data, in square millimeters is shown below:
0 260 356 403 536 0 268 369 428 536
268 396 469 536 162 338 403 536 536 130
A) Construct a 95% confidence interval for the true mean.
B) Why might normality be an issue here?
C) What sample size would be needed to obtain an error of ± 10 square millimeters with 99% confidence?
D) If this is not a reasonable requirement, suggest one that is.
A)
Number of cases =20
Mean or Xbar = 346.5
Squared deviations=
(0-346.5)²= (-346.5)²= 120062.25
(260-346.5)²= (-86.5)² = 7482.25
(356-346.5)² = (9.5)² = 90.25
(403-346.5)² = (56.5)² = 3192.25
(536-346.5)² = (189.5)² = 35910.25
(0-346.5)² = (-346.5)² =120062.25
(268-346.5)²= (-78.5)² =6162.25
(369-346.5)² = (22.5)² =506.25
(428-346.5)² = (81.5)² = 6642.25
(536-346.5)² = (189.5)² =35910.25
(268-346.5)² = (-78.5)² = 6162.25
(396-346.5)² = (49.5)² =2450.25
(469-346.5)² = (122.5)²=15006.25
(536-346.5)² = (189.5)² =35910.25
(162-346.5)²= (-184.5)²=34040.25
(338-346.5)² = (-8.5)²= 72.25
(403-346.5)²= (56.5)²=3192.25
(536-346.5)²= (189.5)²=35910.25
(536-346.5)²= (189.5)² = 35910.25
(130-346.5)²= (-216.5)² =46872.25
Variance = 551547/19
Variance= 29028.7895
Standard deviation =sqrt (variance)
=sqrt (551547/19)
S= 170.3784
Xbar (sample mean) = 346.5
N=20
Se=(standard error)= sd/sqrt(n)= 38.0978
t= 2.093 (refers to the t-distribution critical value with 19 degrees of freedom)
xbar-t se=346.5 –(2.093)(38.0978)=
346.5-79.739=266.76
Xbar +t se= 346.5 + (2.093)(38.0978)=
346.5 + 79.739=426.24
266.76 <µ<426.24
266.76<346.5<426.24
B)
There is a small sample and the data is not continuous since there are two zeros. The normality can be tested by entering the data and frequency to see if the data is normal. Zero has a frequency of 2.536
C)
N=(2.57 sd/10)²=1917 samples
2.57 corresponds to 99% confidence assuming normality.
D)
99% confidence is okay assuming normality. If the distribution is not normal, reducing it to 90% confidence would reduce the required sample size.