Question

a random sample of 828 births included 427 boys. use a 0.01 significance level to test the claim that 50.9% of newborn babies are boys. do the results support the belief that 50.9% of newborn babies are boys? do the results support the belief that 50.9% of newborn babies are boys? a. the results support the belief that 50.9% of newborn babies are boys because there was no evidence to show that the belief is untrue. b. the results do not support the belief that 50.9% of newborn babies are boys because there was sufficient evidence to show that the belief is untrue. c. the results support the belief that 50.9% of newborn babies are boys because there was sufficient evidence to show that the belief is true. d. the results do not support the belief that 50.9% of newborn babies are boys; the results merely show that there is not strong evidence against the rate of 50.9%.

Explanation:

Step1: Calculate sample proportion

The sample proportion \(\hat{p}=\frac{427}{828}\approx0.516\)

Step2: State hypotheses

\(H_0:p = 0.509\), \(H_1:p
eq0.509\)

Step3: Calculate test statistic

The formula for the test statistic \(z\) in a proportion test is \(z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\)
Substitute \(\hat{p}=0.516\), \(p = 0.509\), \(n = 828\)
\(z=\frac{0.516 - 0.509}{\sqrt{\frac{0.509\times(1 - 0.509)}{828}}}\approx0.40\)

Step4: Find critical value and p - value

For a two - tailed test with \(\alpha=0.01\), the critical values are \(z=\pm2.58\)
The p - value for \(z = 0.40\) (two - tailed) is \(P(|Z|>0.40)=2\times(1 - P(Z\leq0.40))\)
From the standard normal table, \(P(Z\leq0.40)=0.6554\), so \(p - value=2\times(1 - 0.6554)=0.6892\)

Step5: Make decision

Since \(|z|=0.40<2.58\) (or \(p - value = 0.6892>0.01\)), we fail to reject \(H_0\)

Failing to reject the null hypothesis does not mean we have proven the null hypothesis is true. It just means we do not have strong evidence against it.

Answer:

D. The results do not support the belief that 50.9% of newborn babies are boys; the results merely show that there is not strong evidence against the rate of 50.9%.