Question

In a study of pregnant women and their ability to correctly predict the sex of their​ baby, 58 of the pregnant women had 12 years of education or​ less, and 41.4​% 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?

Identify the null and alternative hypotheses. Choose the correct answer below. A. H0​: pequals0.5 H1​: pnot equals0.5 B. H0​: pequals0.414 H1​: pless than0.414 C. H0​: pequals0.5 H1​: pgreater than0.5 D. H0​: pequals0.414 H1​: pnot equals0.414 E. H0​: pequals0.414 H1​: pgreater than0.414 F. H0​: pequals0.5 H1​: pless than0.5

The test statistic is zequals nothing. ​(Round to two decimal places as​ needed.)

The​ P-value is nothing. ​(Round to four decimal places as​ needed.)

Identify the conclusion about the null hypothesis. Do the results suggest that their percentage of correct predictions is different from results expected with random​ guesses? ▼ Reject Fail or to reject H0. There ▼ is or not 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 is very different from results expected with random guesses.

Answer 1

Ho :   p =    0.5
H1 :   p ╪   0.5

Level of Significance,   α =    0.01                  
Number of Items of Interest,   x =   24.012                  
Sample Size,   n =    58                  
                          
Sample Proportion ,    p̂ = x/n =    0.4140                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0657                  
Z Test Statistic = ( p̂-p)/SE = (   0.4140   -   0.5   ) /   0.0657 =   -1.31

p-Value   =   0.1902 [excel formula =2*NORMSDIST(z)]  

p value>α ,do not reject null hypothesis   

▼ Fail to reject H0. There ▼ is not 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  very different from results expected with random guesses.