In a study of pregnant women and their ability to correctly predict the sex of t

Question

In a study of pregnant women and their ability to correctly predict the sex of their baby, 56 of the pregnant women had 12 years of education or less, and 44.6% of these women correctly predicted the sex of their baby. Use a 0.01 significance level to test the claim that these women have an ability to predict the sex of their baby equivalent to random guesses. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, and conclusion about the null hypothesis. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution. Do the results suggest that their percentage of correct predictions is different from results expected with random guesses?

a) The test statistic is z=

b)The P-value is=

c) Identify the conclusion about the null hypothesis. Do the results suggest that their percentage of correct predictions is different from results expected with randomguesses? Fill in the spaces below:

___ Reject or Fail to Reject?______ H0. There ___is not or is?_____ sufficient evidence to warrant rejection of the claim that these women have an ability to predict the sex of their baby equivalent to random guesses. The results for these women with 12 years of education or less suggests that their percentage of correct predictions __is not or there is?______ very different from results expected with random guesses.

Explanation / Answer

Given that,
possibile chances (x)=24.976
sample size(n)=56
success rate ( p )= x/n = 0.446
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=0.446-0.5/(sqrt(0.25)/56)
zo =-0.8082
| zo | =0.8082
critical value
the value of |z | at los 0.01% is 2.58
we got |zo| =0.808 & | 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 != -0.8082 ) = 0.41898
hence value of p0.01 < 0.419,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p!=0.5
a.
test statistic: -0.8082
critical value: -2.58 , 2.58
b.
p-value: 0.41898 =41.898%
c.
decision: do not reject Ho
we do not have enough evidence to support the claim that these women have an ability to predict the sex of their baby equivalent to random guesses. The results for these women with 12 years of education or less suggests that their percentage of correct predictions very different from results expected with random guesses.

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