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a random sample of 859 births included 426 boys. use a 0.01 significanc…

Question

a random sample of 859 births included 426 boys. use a 0.01 significance level to test the claim that 50.6% of newborn babies are boys. do the results support the belief that 50.6% of newborn babies are boys?

h₀: p = 0.506

identify the test statistic for this hypothesis test.

the test statistic for this hypothesis test is
(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.)

identify the conclusion for this hypothesis test.

a. fail to reject h₀. there is sufficient evidence to warrant rejection of the claim that 50.6% of newborn babies are boys

Explanation:

Step1: Calculate the sample proportion

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 426$ (number of boys) and $n=859$ (sample size).
$\hat{p}=\frac{426}{859}\approx0.496$

Step2: Calculate the test - statistic

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 $p = 0.506$ (the hypothesized proportion).

$$ LATEXBLOCK0 $$

Step3: Calculate the P - value

Since this is a two - tailed test ($H_1:p
eq0.506$), the P - value is $2\times P(Z\lt|z|)$.
Using a standard normal table or calculator, $P(Z\lt0.58)\approx0.7190$. So the P - value $=2\times(1 - 0.7190)=2\times0.281 = 0.562$

Answer:

The test statistic for this hypothesis test is $-0.58$.
The P - value for this hypothesis test is $0.562$.
For the conclusion:
Since the P - value ($0.562$) $>0.01$ (significance level $\alpha$), we fail to reject $H_0$.
So the answer is:

  • Test statistic: $-0.58$
  • P - value: $0.562$
  • Conclusion: Fail to reject $H_0$. There is not sufficient evidence to warrant rejection of the claim that $50.6\%$ of newborn babies are boys.