Question

1. in a study conducted recently 675 out of 1070 people surveyed expressesed preference for choclate ice cream over other flavors. Let p be the proportion of the population surveyed that prefers choclate ice cream to other flavors.

a. at 5% significance level does the data presents evidence that more than 60% or people from the surveyed population prefer choclate ice cream to other flavors? show both your critical values and p-value hypothesis tests.

b. at 10% significance level does the data presents evidence that the percent of the popluation who prefers choclate ice cream differs from 60%? show both the critical values and p-value hypothesis test.

c. obtain a 90% confidence interval for p

d. based on your Confidence interval, do you think p>.60? explain why or why not

e. what sample size is needed to cut margin or error in your interval from part (c) to at most 1.5%? use 63% as a best guess for p̂

Answer 1

1.

a.

Given that,

possibile chances (x)=675

sample size(n)=1070

success rate ( p )= x/n = 0.631

success probability,( po )=0.6

failure probability,( qo) = 0.4

null, Ho:p=0.6  

alternate, H1: p>0.6

level of significance, alpha = 0.05

from standard normal table,right tailed z alpha/2 =1.645

since our test is right-tailed

reject Ho, if zo > 1.645

we use test statistic z proportion = p-po/sqrt(poqo/n)

zo=0.63084-0.6/(sqrt(0.24)/1070)

zo =2.059

| zo | =2.059

critical value

the value of |z alpha| at los 0.05% is 1.645

we got |zo| =2.059 & | z alpha | =1.645

make decision

hence value of | zo | > | z alpha| and here we reject Ho

p-value: right tail - Ha : ( p > 2.05928 ) = 0.01973

hence value of p0.05 > 0.01973,here we reject Ho

---------------

null, Ho:p=0.6

alternate, H1: p>0.6

test statistic: 2.059

critical value: 1.645

decision: reject Ho

p-value: 0.01973

we have enough evidence to support the claim that more than 60% or people from the surveyed

population prefer choclate ice cream to other flavors.

b.

Given that,

possibile chances (x)=675

sample size(n)=1070

success rate ( p )= x/n = 0.631

success probability,( po )=0.6

failure probability,( qo) = 0.4

null, Ho:p=0.6  

alternate, H1: p!=0.6

level of significance, alpha = 0.1

from standard normal table, two tailed z alpha/2 =1.645

since our test is two-tailed

reject Ho, if zo < -1.645 OR if zo > 1.645

we use test statistic z proportion = p-po/sqrt(poqo/n)

zo=0.63084-0.6/(sqrt(0.24)/1070)

zo =2.059

| zo | =2.059

critical value

the value of |z alpha| at los 0.1% is 1.645

we got |zo| =2.059 & | z alpha | =1.645

make decision

hence value of | zo | > | z alpha| and here we reject Ho

p-value: two tailed ( double the one tail ) - Ha : ( p != 2.05928 ) = 0.03947

hence value of p0.1 > 0.0395,here we reject Ho

---------------

null, Ho:p=0.6

alternate, H1: p!=0.6

test statistic: 2.059

critical value: -1.645 , 1.645

decision: reject Ho

p-value: 0.03947

we have enough evidence to support the claim that the percent of the popluation who prefers choclate ice cream differs from 60%

c.

TRADITIONAL METHOD

given that,

possibile chances (x)=675

sample size(n)=1070

success rate ( p )= x/n = 0.631

I.

sample proportion = 0.631

standard error = Sqrt ( (0.631*0.369) /1070) )

= 0.015

II.

margin of error = Z a/2 * (stanadard error)

where,

Za/2 = Z-table value

level of significance, α = 0.1

from standard normal table, two tailed z α/2 =1.645

margin of error = 1.645 * 0.015

= 0.024

III.

CI = [ p ± margin of error ]

confidence interval = [0.631 ± 0.024]

= [ 0.607 , 0.655]

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DIRECT METHOD

given that,

possibile chances (x)=675

sample size(n)=1070

success rate ( p )= x/n = 0.631

CI = confidence interval

confidence interval = [ 0.631 ± 1.645 * Sqrt ( (0.631*0.369) /1070) ) ]

= [0.631 - 1.645 * Sqrt ( (0.631*0.369) /1070) , 0.631 + 1.645 * Sqrt ( (0.631*0.369) /1070) ]

= [0.607 , 0.655]

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interpretations:

1. We are 90% sure that the interval [ 0.607 , 0.655] contains the true population proportion

2. If a large number of samples are collected, and a confidence interval is created

for each sample, 90% of these intervals will contains the true population proportion

d.

yes,

confidence interval is more than the 60%. because the population proportion more than 60% or people from the surveyed population prefer choclate ice cream to other flavors

e.

Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)

Z a/2 at 0.1 is = 1.645

Sample Proportion = 0.63

ME = 0.015

n = ( 1.645 / 0.015 )^2 * 0.63*0.37

= 2803.442 ~ 2804