please explan without using excel ... 5 answers ... hw 8.3 # 3 In a study of pre

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please explan without using excel ... 5 answers ... hw 8.3 # 3

In a study of pregnant women and their ability to correctly predict the sex of their baby, 57of the pregnant women had 12 years of education or less, and 43.9%o f these women correctly predicted the sex of their baby. Use a 0.05 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?

1.)Identify the null and alternative hypotheses. Choose the correct answer below.

A.H0:pequals=0.50.5H1:pnot equals0.50.5
B.H0:pequals=0.4390.439H1:pnot equals0.4390.439
C.H0:pequals=0.4390.439H1:pless than<0.4390.439
D.H0:pequals=0.50.5H1:pless than<0.50.5
E.H0:pequals=0.4390.439H1:pgreater than>0.4390.439
F.H0:pequals=0.50.5H1:pgreater than>0.50.5

2) The test statistic is z= _________(Round to 2 decimal places as needed.)
3)The P-value is _________ (Round to 4 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?

(3)_________? H0.
(4) There___________?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
(5)________?very different from results expected with random guesses.

(3)Fail to reject/Reject
(4)is/is not
(5)is/is not

Explanation / Answer

Identify the null and alternative hypotheses.

Null hypothesis: p=0.5

Alternative hypothesis: p not equal to 0.5

option A) is correct

The test statistic is z=(phat-p)/sqrt(p*(1-p)/n)

=(0.439-0.5)/sqrt(0.5*0.5/57)

= -0.921079

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The p-value is 2*P(Z<-0.921079)= 2*0.1785 = 0.357 (from standard normal table)

Identify the conclusion about the null hypothesis and the final conclusion that addresses the original claim.

Since the p-value is larger than 0.05, we do not reject the null hypothesis.

So we can conclude that these women have no ability to predict the sex of their baby, and the results are not significantly different from those that would be expected with random guesses.

hence

1)option A)

2) -0.921079

3)0.357

4) fail to reject Ho

5) not suffiecient evidence

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