The confidence interval for the proportion of students attending a football game is [0.6592, 0.7044].
Confidence interval can be calculated by below formula

where the left part is the lower interval and right side is the higher interval.
Pcap = 750/1100
Pcap = 0.6818
Again to calculate
, as we know that the level of confidence is 90 percent so its value will be 1.645
now putting the values on the both side we can get ,
0.6818 - 1.645 x 0.014 < p < 0.6818 + 1.645 x 0.014
0.6818 - 0.023 < p < 0.6818 + 0.023
0.6588 < p < 0.7048
the nearest value is [0.6592, 0.7044]
Hence the confidence interval for the proportion of students attending a football game is [0.6592, 0.7044]
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Answer:
x =2
Step-by-step explanation:
x + 7 = 6x - 3
7 + 3 = 6x - x
10 = 5x
x = 10 ÷ 5
x = 2
725,278 * 67,066 = 48641494348....now, if x is in the 10 digits spot...it is 4...and if y is in the units spot, then it is 8.....
x + y = 4 + 8 = 12
Answer:
a)
<em> you are unwilling to predict the proportion value at your school = 0.90</em>
<em>b) </em>
<em>The large sample size 'n' = 864</em>
<em></em>
Step-by-step explanation:
Given estimated proportion 'p' = 10% = 0.10
<em>Given Margin of error M.E = 0.02</em>
<em>Level of significance α = 0.05</em>
a)
<em> you are unwilling to predict the proportion value at your school </em>
<em> q = 1- p = 1- 0.10 =0.90</em>
b)
<em>The Margin of error is determined by</em>


Cross multiplication , we get

√n = 29.4
<em>Squaring on both sides , we get</em>
<em> n = 864.36</em>
<u><em>Conclusion</em></u><em>:-</em>
<em>The large sample size 'n' = 864</em>
<em></em>
<em></em>
Answer:
Step-by-step explanation:
you do 11\3 divided by 1\5