the students receive. A recent poll asked 1,175 parents who have children Severa
ID: 3311660 • Letter: T
Question
the students receive. A recent poll asked 1,175 parents who have children Several years ago 41% of parents who had chldren n grades K-12 were satisfied with the quality of education in grades K-12 if they were satisfied with the quality of education the students receive. Of the 1,175 surveyed, 473 indicated that they were satisfied C to assess whether this represents evidence that parents attitudes toward the quality of education have changed What are the null and altermative hypotheses? (Round to two decimal places as needed)Explanation / Answer
Ho: P = 0.41, H1: P != 0.41
TRADITIONAL METHOD
given that,
possibile chances (x)=473
sample size(n)=1175
success rate ( p )= x/n = 0.4026
I.
sample proportion = 0.4026
standard error = Sqrt ( (0.4026*0.5974) /1175) )
= 0.0143
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.0143
= 0.028
III.
CI = [ p ± margin of error ]
confidence interval = [0.4026 ± 0.028]
= [ 0.3745 , 0.4306]
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DIRECT METHOD
given that,
possibile chances (x)=473
sample size(n)=1175
success rate ( p )= x/n = 0.4026
CI = confidence interval
confidence interval = [ 0.4026 ± 1.96 * Sqrt ( (0.4026*0.5974) /1175) ) ]
= [0.4026 - 1.96 * Sqrt ( (0.4026*0.5974) /1175) , 0.4026 + 1.96 * Sqrt ( (0.4026*0.5974) /1175) ]
= [0.3745 , 0.4306]
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interpretations:
1. We are 95% sure that the interval [ 0.3745 , 0.4306] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
90% CI
confidence interval = [ 0.4026 ± 1.645 * Sqrt ( (0.4026*0.5974) /1175) ) ]
= [0.4026 - 1.645 * Sqrt ( (0.4026*0.5974) /1175) , 0.4026 + 1.645 * Sqrt ( (0.4026*0.5974) /1175) ]
= [0.38 , 0.43]
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