Use the sample data and confidence level given below to complete parts (a) through (d). In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2607 subjects randomly selected from an online group involved with ears. 1019 surveys were returned. Construct a 90% confidence interval for the proportion of returned surveys. E Click the icon to view table of z scores. a) Find the best point estimate of the population proportion p. (Round three decimal places as needed.) b) Identify the value of the margin of error E. E=O (Round to three decimal places as needed.) c) Construct the confidence interval. (Round to three decimal places as needed.) d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below. O A. There is a 90% chance that the true value of the population proportion will fall between the lower bound and the upper bound. O B. 90% of sample proportions will fall between the lower bound and the upper bound. Oc. One has 90% confidence that the sample proportion is equal to the population proportion. O D. One has 90% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

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Use the sample data and confidence level given below to complete parts (a) through (d).
In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2607 subjects randomly selected from an online group involved with ears. 1019 surveys were returned. Construct a 90% confidence interval for
the proportion of returned surveys.
E Click the icon to view a table of z scores.
a) Find the best point estimate of the population proportion p.
(Round to three decimal places as needed.)
b) Identify the value of the margin of error E.
E=
(Round to three decimal places as needed.)
c) Construct the confidence interval.
]<p<[
(Round to three decimal places as needed.)
d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below.
O A. There is a 90% chance that the true value of the population proportion will fall between the lower bound and the upper bound.
B. 90% of sample proportions will fall between the lower bound and the upper bound.
One has 90% confidence that the sample proportion is equal to the population proportion.
O D. One has 90% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.
Transcribed Image Text:Use the sample data and confidence level given below to complete parts (a) through (d). In a study of cell phone use and brain hemispheric dominance, an Internet survey was e-mailed to 2607 subjects randomly selected from an online group involved with ears. 1019 surveys were returned. Construct a 90% confidence interval for the proportion of returned surveys. E Click the icon to view a table of z scores. a) Find the best point estimate of the population proportion p. (Round to three decimal places as needed.) b) Identify the value of the margin of error E. E= (Round to three decimal places as needed.) c) Construct the confidence interval. ]<p<[ (Round to three decimal places as needed.) d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below. O A. There is a 90% chance that the true value of the population proportion will fall between the lower bound and the upper bound. B. 90% of sample proportions will fall between the lower bound and the upper bound. One has 90% confidence that the sample proportion is equal to the population proportion. O D. One has 90% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.
POSITIVE z Scores
Standard Normal (2) Distribution: Cumulative Area from the LEFT
00
00
5000
S040 so0 5120 S160
00
01
0.1
02
5790
6103
02
03
8179
406
6517
03
04
ST00
04
0.5
915
7e8 712
724
0.5
0.6
744
07
7580
2611 J642 3
07
08
0.9
8159
212 230
8264
|時
09
1.0
8413
4
1.0
1.1
AT29
770
A790
1.1
12
12
1.3
131
13
14
14
1.5
se
441
1.5
43 474
s : 0s
645
1.4
1.6
9 08 16 25 33
1.7
9554
91
17
18
41
78
18
S719 2 S732
1.9
S713
9738
744
1.9
20
20
21
21
2 3M
042
21
22
71
78
22
23
04
011
23
24
es
931
32
24
25
25
26
9960
963
26
27
970
071
27
28
974
79
9980
81
28
29
e se
29
30
987
3.0
31
31
32
94
32
33
33
34
34
3.50. nd up
3.50. nd up
00
04
07
08
Standard Normal (z) Distribution: Cumulative Area from the LEFT
Transcribed Image Text:POSITIVE z Scores Standard Normal (2) Distribution: Cumulative Area from the LEFT 00 00 5000 S040 so0 5120 S160 00 01 0.1 02 5790 6103 02 03 8179 406 6517 03 04 ST00 04 0.5 915 7e8 712 724 0.5 0.6 744 07 7580 2611 J642 3 07 08 0.9 8159 212 230 8264 |時 09 1.0 8413 4 1.0 1.1 AT29 770 A790 1.1 12 12 1.3 131 13 14 14 1.5 se 441 1.5 43 474 s : 0s 645 1.4 1.6 9 08 16 25 33 1.7 9554 91 17 18 41 78 18 S719 2 S732 1.9 S713 9738 744 1.9 20 20 21 21 2 3M 042 21 22 71 78 22 23 04 011 23 24 es 931 32 24 25 25 26 9960 963 26 27 970 071 27 28 974 79 9980 81 28 29 e se 29 30 987 3.0 31 31 32 94 32 33 33 34 34 3.50. nd up 3.50. nd up 00 04 07 08 Standard Normal (z) Distribution: Cumulative Area from the LEFT
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