Test the claim that 11% of a candy maker\'s candies are red Use a 0 05 significa
ID: 3315570 • Letter: T
Question
Test the claim that 11% of a candy maker's candies are red Use a 0 05 significance level. Use the data in the table to the rightto answer the following questions Weights (g) of a Sample Bag of Candy Red 0.902 0.715 0.921 0954 0.953 0.916 0.976 0.787 Blue 0.986 944 0.892 0.907 0.701 0.927 0.924 Yellow 0.868 0.811 0.726 0.846 0.787 0.998 0.958 0.796 0946 Green a. Find the sample proportion of candies that are red red = (Round to three decimal places as needed.) b. Which of the following is the hypothesis test to be conducted? OA. Ho:p=0.11 Brown 0.928 0.701 0.815 0.866 0.923 0.938 0.933 0.768 0.809 0.956 0.786 0.982 0.815 0.723 0 792 0.826 0.927 0.963 0.977 O B. Ho p 011 H1 p>0.11 O C. Ho pc0.11 D. My p=0.11 Ho:p>011 H p 0.11 Ho p#011 HI p=0.11 Hop:0.11 Hip#011 OE. F. c. What is the test statistic? Click to select your answer(s)Explanation / Answer
Given that,
red = 8
blue = 7
brown = 10
green = 11
yellow = 9
possible chances (x)=8
sample size(n)=45
success rate ( p )= x/n = 0.1778
success probability,( po )=0.11
failure probability,( qo) = 0.89
null, Ho:p=0.11
alternate, H1: p!=0.11
level of significance, = 0.05
from standard normal table, two tailed z /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=0.17778-0.11/(sqrt(0.0979)/45)
zo =1.4531
| zo | =1.4531
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =1.453 & | z | =1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.45312 ) = 0.14619
hence value of p0.05 < 0.1462,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.11
alternate, H1: p!=0.11
test statistic: 1.45
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.14619
Option A, suffcient evidence
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