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20) A clinical trial was conducted using a new method designed to increase
the probability of conceiving a girl. As of this writing, 914 babies were
born to parents using the new method, and 877 of them were girls. Use a
(b).01 significance level to test the claim that the new method is effective in
increasing the likelihood that a baby will be a girl. Use the P-value method
and the normal distribution as an approximation to the binomial
distribution.
(10)
a. Identify the null and alternative hypothesis.
b. Compute the test statistic z.ollowenien e to jaata ni(8)
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c. What is the P-value?
d. What is the conclusion about the null hypothesis?
e. What is the final conclusion?

The test statistic is 27.8484, the p-value is 0, and the final conclusion is that the null hypothesis H is rejected.

Given that 914 babies were born to parents who used the new method, and 877 of them were girls, the significance level to test the claim that the new method is effective in increasing the likelihood of a baby being a girl, is 0.01.

The following information is provided: the sample size is N = 914, the number of favorable cases is X = 877, and the sampling ratio is

pˉ = X / N

Pˉ = 877/914

pˉ = 0.9595 and the significance level is α = 0.01

(a) Hypothesis Zero and Alternative

The following null and alternative hypotheses should be tested:

null: p = 0.5

Alternative: p> 0.5

This is equivalent to a right-tailed test, which requires a z-test for a proportion of the population.

(b) Critical Value

Based on the information provided, the significance level is α = 0.01, so the critical value for this right-tailed test is Zc = 2.3263. This can be found using Excel or the Z distribution table. Region of rejection

The rejection area for this test on the right side is Z> 2.3263

Test statistics

The z-statistic is calculated as follows:

[tex]\begin{aligned}Z&=\frac{\bar{p}-p}{\sqrt{\frac{p(1-p)}{n}}}\\ &=\frac{0.9595-0.50}{\sqrt{\frac{0.5(1-0.5)}{914}}}\\ &=\frac{0.4595}{0.0165}\\ &=27.8484\end[/tex]

(c) The p-value

The p-value is the probability that the sample results are extreme or more extreme than the sample results obtained, assuming the null hypothesis is true. In this case,

the p-value is p = P (Z> 27.8484) = 0

(d) The decision on the null hypothesis

Using the traditional method

Since we observe that Z = 27.8484> Zc = 2.3263, we conclude that the null hypothesis is rejected.

Using the p-value method

Using the approximation of the P-value: the p-value is p = 0, and since p = 0≤0.01 we conclude that the null hypothesis is rejected.

(e) Conclusion

It is concluded that the null hypothesis H is rejected. Therefore, there is sufficient evidence to state that the population proportion p is greater than 0.5, at the significance level of 0.01.

Hence, for 914 babies born to parents using the new method, of which 877 were girls, the test statistic is 27.8484, the p-value is 0, and the final conclusion is that the null hypothesis is rejected.

Learn more about test statistics from here brainly.com/question/14805123

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Megan’s TV had a base price of $189 and the sales tax was 6.88%. Over the next four years, the TV consumed an average of $0.06 of electricity every day. It also needed repairs twice, costing $29 each time. After four years, Megan got a new TV. What was the lifetime cost of Megan’s TV? (Remember that one in every four years is a leap year.)

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