A survey of an urban university showed that 800 of 1000 students sampled support

Question

A survey of an urban university showed that 800 of 1000 students sampled supported a fee increase to fund improvements to the student recreation center. 3. a. What is the point estimate of the population proportion ( b. What is the standard error of the sample proportion (p)? c. What is the probability that sample proportion (p) is less than 0.7? d. Using the 99% level of confidence, what is the confidence interval of ? e, The President will increase the fee if 85% of the students support the fee increases. What will the President do? )?

Explanation / Answer

P = 800/1000 = 0.8 , n= 1000
a) 0.8 is the point estimate

b)

SE = sqrt(p * ( 1 -p)/n)
= sqrt( 0.8 * 0.2 / 1000)
= 0.0126

c)
P(p < 0.7)

z = ( p - P)/ SE
= ( 0.7 -0.8) / 0.0126
= -7.936

P(p<0.7) = P( z < -0.7936) = 0

d) z value at 99% = 2.58

CI = P +/- z *SE
= 0.8 +/- 2.58 * 0.0126
= 0.767 - 0.833

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