Question

Mr Smith is on another tirade against honeydew. He found another fruit bowl which had 18 out of 45 honeydew. He once again believes this is an affront to his rule which says honeydew must be at 30%. This had more! Does he have a right to believe the store is increasing the amount .  The standard error for his study will be .0683.

a) (4 points) Create the Hypothesis Test for this scenario.

b) (2 points) Find the test statistic for this study.

c) (2 points) Draw a sketch (or use Statkey) for the normal curve (or snip it). Show your center, test statistic, shading and p-value in your sketch

d) (4 points) Make your statistical and practical conclusions for the study.

e) (2 points) We want to create a confidence interval to support this Hypothesis study. To do this, first, find the critical value for the 98% confidence interval.

f)  (3 points) Find the 98% confidence interval. Show your calculations.

g) (3 points) Does the confidence interval support your Hypothesis Test. Why or why not?

Answer 1

Given that,
possibile chances (x)=18
sample size(n)=45
success rate ( p )= x/n = 0.4
success probability,( po )=0.3
failure probability,( qo) = 0.7
null, Ho:p=0.3
alternate, H1: p>0.3
level of significance, α = 0.02
from standard normal table,right tailed z α/2 =2.054
since our test is right-tailed
reject Ho, if zo > 2.054
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.4-0.3/(sqrt(0.21)/45)
zo =1.464
| zo | =1.464
critical value
the value of |z α| at los 0.02% is 2.054
we got |zo| =1.464 & | z α | =2.054
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: right tail - Ha : ( p > 1.46385 ) = 0.07162
hence value of p0.02 < 0.07162,here we do not reject Ho
ANSWERS
---------------
a.
null, Ho:p=0.3
alternate, H1: p>0.3
b.
test statistic: 1.464
decision: do not reject Ho
c.
p-value: 0.07162
p value is greater than alpha value
shaded region in the right sided in the figure
d.
we do not have enough evidence to support the claim that right to believe the store is increasing the amount .
e.
critical value: 2.054
f.
TRADITIONAL METHOD
given that,
possible chances (x)=18
sample size(n)=45
success rate ( p )= x/n = 0.4
I.
sample proportion = 0.4
standard error = Sqrt ( (0.4*0.6) /45) )
= 0.073
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.326
margin of error = 2.326 * 0.073
= 0.17
III.
CI = [ p ± margin of error ]
confidence interval = [0.4 ± 0.17]
= [ 0.23 , 0.57]
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DIRECT METHOD
given that,
possible chances (x)=18
sample size(n)=45
success rate ( p )= x/n = 0.4
CI = confidence interval
confidence interval = [ 0.4 ± 2.326 * Sqrt ( (0.4*0.6) /45) ) ]
= [0.4 - 2.326 * Sqrt ( (0.4*0.6) /45) , 0.4 + 2.326 * Sqrt ( (0.4*0.6) /45) ]
= [0.23 , 0.57]
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interpretations:
1. We are 98% sure that the interval [ 0.23 , 0.57] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population proportion
g.
No,
the confidence interval supports the hypothesis test because,
it fails to reject the null hypothesis.