Chapter 9.2, Problem 45E

(a)

To determine

To find: The population value in which the study being generalized.

Answer to Problem 45E

The collected sample will represent the population of firstborn children.

Explanation of Solution

Given information:

A random sample of 25,468 firstborn children included 13,173 boys.13 Boys do make up more than half of the sample.

 Identify the population to which the results of the study can be generalized:

It is given that, a random sample of 25,468 firstborn children included 13,173 boys.

Here, all the sample of children are firstborn children.

Therefore, the collected sample will represent the population of firstborn children.

Hence, the results can be generalized to all firstborn children.

Conclusion:

The collected sample will represent the population of firstborn children.

(b)

To determine

To analyse: The boys are more common than girls.

Answer to Problem 45E

There is enough evidence to support the claim that the boys are more common than girls.

Explanation of Solution

Given information:

A random sample of 25,468 firstborn children included 13,173 boys.13 Boys do make up more than half of the sample.

Test whether boys are more common than girls in the population:

The investigator is specifically interested to test whether boys are more common than girls in the population.

Denote the population proportion of boys as p.

Denote the sample proportion of boys as p^.

Null hypothesis:

H₀: p  = 0.5

That is, the boys are not more common than girls.

Alternative hypothesis:

H₁: p  > 0.5

That is, the boys are more common than girls.

Sample proportion:

It is given that among a random sample of 25,468 firstborn children, 13173 are boys.

The sample size is $n =25468$ .

The number of specified characteristics is $x =13173$ .

The sample proportion of boys is 

  $\begin{array}{l}\text{p}= \frac{x}{n}\\ =\frac{13173}{25468}\\ \text{p}=\mathrm{0.5172.}\end{array}$

Test statistic:

The test statistic is obtained as 5.5484 from the calculations given below:

  $\begin{array}{l}z=\frac{\hat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}\\ =\frac{0.5172-0.5}{\sqrt{\frac{0.5\left(1-0.5\right)}{25,468}}}\\ =\frac{0.0172}{0.0031}\\ z=5.5484\end{array}$

Thus, the test statistic value is 5.5484.

P-value:

The test statistic is z = 5.5484.

The hypothesis test is right tailed.

The P-value is obtained as 0 from the calculation given below:

  $\begin{array}{l}P-\text{value}=P_{H_0}\left(z>5.5484\right)\\ =1-P_{H_0}\left(z<5.5484\right)\\ =1-\Phi \left(5.5484\right)\\ =1-1\left[\begin{array}{l}\text{From Z-table the cumulative probalitity at the intersection }\\ \text{of the row value of z 5}\text{.5 and column value of z 0}\text{.05}\\ \because \text{EXCEL}\text{FORMULA:}\\ \text{"NORM}\text{.S}\text{.DIST}\left(5.5484,\text{TRUE}\right)"\end{array}\right]\\ P-\text{value}=0\end{array}$

Thus, the  P -value is 0.

Since, the level of significance is not specified, the prior level of significance a = 0.05 can be used.

Decision rule based on P-value approach:

If P-value = a, then reject the null hypothesis H₀.

If P-value > a, then fail to reject the null hypothesis H₀.

Conclusion based on P-value approach:

The P-value is 0 and a value is 0.05.

Here, P-value is less than the a value.

That is, 0 (=P-value) < 0.05 (=a).

By the rejection rule, reject the null hypothesis.

Thus, there is enough evidence to support the claim that the boys are more common than girls.

Conclusion:

There is enough evidence to support the claim that the boys are more common than girls.