PLEASE HELP SOLVE. PLEASE EXPLAIN 1. 2. 3. 1) For a particular study, 599 parent

Question

PLEASE HELP SOLVE. PLEASE EXPLAIN

1.

2.

3.

1) For a particular study, 599 parents were asked if they needed a permanent babysitter. 76% of them said yes. What is the approximate Margin of Error for this poll? Margin of Error (Round your answer to 2 decimal places) 2) At another location and asking the same question, the margin of error was ±2.14% and 79% said yes. a.) Approximately how many parents did they poll for this study? parents (Round to the nearest whole pareent) b.) What is the confidence interval for "Yes?" 95%-confidence interval-

Explanation / Answer

1)
p = 0.76 , n = 599

assume z value at 95% Ci = 1.96

Margin of error = z * sqrt(p *(1 - p) / n )
= 1.96 * sqrt( 0.76 * ( 1 - 0.76) / 599)
= 0.0342 = 3.42%

2)

a) Confidence interval is not given, so assume z value at 95% CI = 1.96

margin of error = +/-2.14 , p = 0.79

Margin of error = z * sqrt(p * ( 1 -p)/n)

n = (z / margin of error)^2 * (p * ( 1-p))

n = ( 1.96/0.0214)^2 * ( 0.79 * 0.21)

n = 1391.65 = 1392

b)

95% CI = p + / - z * sqrt(p * ( 1 -p) / n)
  
= 0.79 +/- 1.96 * sqrt( 0.79 * 0.21/1392)
= 0.7686 - 0.8114

= 76.86% - 81.14%

3)
13 , 2 , 6 , 9 , 14 , 11 , 12 , 15 , 7 , 3

a) 92 Total people were asked

b)

6 , 9 , 20 , 23 , 30 , 33 , 39 , 44 ,50 , 51

mean = 30.5

c)
median = (30+33)/2= 31.5

d)
There is no repeated number. so no mode

e)
total people for 9yr =2
Percentile = 2 /n
= 2 / 92 = 0.022 = 2.2%

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