Question

In a particular year, 68% of online courses taught at a system of community colleges were taught by full-time faculty. To test if 68% also represents a particular state's percent for full-time faculty teaching the online classes, a particular community college from that state was randomly selected for comparison. In that same year, 31 of the 44 online courses at this particular community college were taught by full-time faculty. Conduct a hypothesis test at the 5% level to determine if 68% represents the state in question.

d) State the distribution to use for the test. (Round your standard deviation to four decimal places.)

f) What is the p-value? (Round your answer to four decimal places.)

g) Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

i) Construct a 95% confidence interval for the true proportion. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to four decimal places.)

Answer 1

Given that,
possibile chances (x)=31
sample size(n)=44
success rate ( p )= x/n = 0.7045
success probability,( po )=0.68
failure probability,( qo) = 0.32
null, Ho:p=0.68
alternate, H1: p!=0.68
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.70455-0.68/(sqrt(0.2176)/44)
zo =0.349
| zo | =0.349
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =0.349 & | z α | =1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.34903 ) = 0.72706
hence value of p0.05 < 0.7271,here we do not reject Ho
ANSWERS
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d.
random sample is normally distributed
null, Ho:p=0.68
alternate, H1: p!=0.68
test statistic: 0.349
critical value: -1.96 , 1.96
decision: do not reject Ho
f.
p-value: 0.72706
g.
p value is greater than alpha value
we do not have enough evidence to support the claim that if 68% also represents a particular state's percent for full-time faculty teaching the online classes,
a particular community college from that state was randomly selected for comparison.
i.
TRADITIONAL METHOD
given that,
possible chances (x)=31
sample size(n)=44
success rate ( p )= x/n = 0.7045
I.
sample proportion = 0.7045
standard error = Sqrt ( (0.7045*0.2955) /44) )
= 0.0688
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.0688
= 0.1348
III.
CI = [ p ± margin of error ]
confidence interval = [0.7045 ± 0.1348]
= [ 0.5697 , 0.8394]
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DIRECT METHOD
given that,
possible chances (x)=31
sample size(n)=44
success rate ( p )= x/n = 0.7045
CI = confidence interval
confidence interval = [ 0.7045 ± 1.96 * Sqrt ( (0.7045*0.2955) /44) ) ]
= [0.7045 - 1.96 * Sqrt ( (0.7045*0.2955) /44) , 0.7045 + 1.96 * Sqrt ( (0.7045*0.2955) /44) ]
= [0.5697 , 0.8394]
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interpretations:
1. We are 95% sure that the interval [ 0.5697 , 0.8394] 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