Question

a random sample of 836 births included 430 boys. use a 0.05 significance level to test the claim that 50.5% of newborn babies are boys. do the results support the belief that 50.5% of newborn babies are boys?
identify the null and alternative hypotheses for this test. choose the correct answer below.
a. ( h_{0}: p = 0.505 )
( h_{1}: p
eq0.505 )
b. ( h_{0}: p = 0.505 )
( h_{1}: p < 0.505 )
c. ( h_{0}: p
eq0.505 )
( h_{1}: p = 0.505 )
d. ( h_{0}: p = 0.505 )
( h_{1}: p > 0.505 )
identify the test statistic for this hypothesis test.
the test statistic for this hypothesis test is 52
(round to two decimal places as needed)
identify the p - value for this hypothesis test.
the p - value for this hypothesis test is
(round to three decimal places as needed)

Explanation:

Step1: Calculate the sample proportion

The sample proportion \(\hat{p}=\frac{x}{n}\), where \(x = 430\) (number of boys) and \(n=836\) (sample size). So \(\hat{p}=\frac{430}{836}\approx0.514\)

Step2: Calculate the test statistic \(z\)

The formula for the test statistic in a hypothesis test for a proportion is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\), where \(p = 0.505\) (claim proportion), \(\hat{p}\approx0.514\), and \(n = 836\)

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Step3: Calculate the P - value

Since this is a two - tailed test (\(H_1:p
eq0.505\)), the P - value is \(2\times(1-\Phi(|z|))\), where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution.

For \(z = 0.52\), \(\Phi(0.52)=0.6985\) (from standard normal table)

The P - value \(=2\times(1 - 0.6985)=2\times0.3015 = 0.603\)

Answer:

The P - value for this hypothesis test is \(0.603\)