A professor at a local university designed an experiment to see if someone could

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A professor at a local university designed an experiment to see if someone could identify the color of a candy based on taste alone. Students were blindfolded and then given a red-colored or yellow-colored candy to chew. (Half the students were assigned to receive the red candy and half to receive the yellow candy. The students could not see what color candy they were given.) After chewing, the students were asked to guess the color of the candy based on the flavor. Of the 124 students who participated in the study, 74 correctly identified the color of the candy. The results are shown in the accompanying technology printout. Complete parts a through c below.

&7.6.72 Question Hep * A profassor at a local university designed an experiment to see if someone could idenry the color of a candy based on taste alone. Students were bindfolded and then given a rad-colored or yellow-colored cand to chew. (Half tha shudents wara assignad to recaive the rad candy and half to acsive the yallaw candy Tha students coul net se what caler candy thaywara given ) Ahar chewing, the studants wara asknd to guass tha colar af the candy hasad an the flavor. Of the 124 students who participated in the study. 74 correctly identified the color of the candy. The resus are shown in the accempanying lechnology prinizut. Complete parts a through c below Click thicon to viaw tha technalogy, printout s na relationship betwo eandy flavar what p oectly enty the cola? The proportion would be·(Type an integer or a decimal.) h. Speciy the nul and altamative hypathasns for testing whathar tolor and flavorae rlatad Choosn the camect hypothases belaa and flavar B. D. OR Ho'p#050 vs Ha' p0.50 Ho-p#05D vs. H, p-0.50 C. Ha : p c 0.50 vs. He: p = 0.50 Ho . p> 0.50 vs. Hs, p-0.50 E. C. Carry out the test and give the appropriate concuson at -0.01. Use the p-value of the est shown on the accompanying technology pr not, to make your decision. The p-value is(Type an integer or a decimal.)

Explanation / Answer

Given that,
possibile chances (x)=74
sample size(n)=124
success rate ( p )= x/n = 0.5968
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p!=0.5
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.59677-0.5/(sqrt(0.25)/124)
zo =2.1553
| zo | =2.1553
critical value
the value of |z | at los 0.01% is 2.58
we got |zo| =2.155 & | z | =2.58
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.15526 ) = 0.03114
hence value of p0.01 < 0.0311,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p!=0.5
test statistic: 2.1553
critical value: -2.58 , 2.58
decision: do not reject Ho
p-value: 0.03114

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