Question

if possible please explain how to obtain answers on ti-84 calculator

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 42.9%

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.

The test statistic is

z=

​(Round to two decimal places as​ needed.)

The​ P-value is

​(Round to four decimal places as​ needed.)

Answer 1

(a)

H0: Null Hypothesis: P = 0.5

HA: Alternative Hypothesis: P 0.5

(b)

n = Sample size = 56

p = Sample proportion = 0.429

SE =

Test statistic is:

Z = (p -P)/Se

= (0.429 - 0.5)/0.0668 = - 1.06

(c)
Table of Area Under Standard Normal Curve gives area = 0.3554

So,

The P-value = (0.5 - 0.3554) X 2 = 0.2892

(d) Since P-value = 0.2892 is greater than = 0.01, Fail to reject null hypothesis.

Conclusion:

The data support the claim that these women have an ability to predict the sex of their baby equivalent to random guesses.