A study addressed the issue of whether pregnant women can correctly predict the gender of their baby. Among 104 pregnant​ women, 57 correctly predicted the gender of their baby. If pregnant women have no such​ ability, there is a 0.327 probability of getting such sample results by chance. What do you​ conclude?

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Use the given probability value to determine whether the sample results could easily occur by​ chance, then form a conclusion.
 
A study addressed the issue of whether pregnant women can correctly predict the gender of their baby. Among 104 pregnant​ women, 57 correctly predicted the gender of their baby. If pregnant women have no such​ ability, there is a 0.327 probability of getting such sample results by chance. What do you​ conclude?
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The following information is provided: The sample size is N=104N = 104, the number of favorable cases is X=57X = 57, and the sample proportion is pˉ=XN=57104=0.5481\bar p = \frac{X}{N} = \frac{ 57}{ 104} = 0.5481, and the significance level is α=0.05\alpha = 0.05

 Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:p=0.327Ho: p = 0.327 (whether pregnant women can correctly predict the gender of their baby IS BY CHANCE)

This corresponds to a two-tailed test, for which a z-test for one population proportion needs to be used.

Rejection Region

Based on the information provided, the significance level is α=0.05\alpha = 0.05 , and the critical value for a two-tailed test is zc=1.96z_c = 1.96.

The rejection region for this two-tailed test is R={z:z>1.96}

 

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