Question

According to an article in Newsweek, the natural ratio of girls to boys is $100:105$. In China, the birth ratio is $100:114$(46.7% girls). Suppose you don’t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in $150$randomly chosen recent births. There are $60$girls and $90$boys born of the $150$. Based on your study, do you believe that the percent of girls born in China is$46.7$?

Short answer

The $95\%$confidence level shows that the proportion of Chinese girls born falls between $32.16\%$and $47.84\%$.

Step-by-step solution

Given information

A hypothesis is a reasonable assumption for a behaviour (plural hypotheses). The scholarly approach involves that an assumption be validated before it can be deemed a factual prediction. Technical speculations are mainly influenced by past results that cannot be fully addressed by established scientific findings.

Explanation

Let's start by deciding on the null and alternate hypotheses:

The null hypothesis indicates that the proportion of Chinese girls born is $46.7\%$, while the alternative hypothesis states that the proportion of Chinese girls born is not $46.7$percent.

$$\begin{gathered} H_0:p=46.7\% \\ {H}_a:p\ne 46.7\% \end{gathered}$$

The random variable here is the proportion of Chinese girls born. As a result, for this test, we'll utilise the normal distribution.

$N\left(0.467,\sqrt{\frac{(0.467)(0.533)}{150}}\right)$

Then the $z-$test is

$z=\frac{p^{\text{'}}-p}{\sqrt{\frac{p(1-p)}{n}}}$

Here, $p\text{'}$is the observed proportion, and we calculate $p\text{'}$using the following method, where n is the sample size of $50$people:

$p\text{'}=\frac{x}{n}$

$$\begin{gathered} =\frac{60}{150} \\ =0.4 \end{gathered}$$

Substitute the $p\text{'}$value

$z=\frac{p^{\text{'}}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$$\begin{gathered} =\frac{-0.067}{0.04074} \\ =-1.645 \end{gathered}$$

Let's use the following formula to determine the p-value for a two-tailed test:

$$\begin{gathered} P=2\times P\left(z>/z_{\text{stat}}/\right) \\ P=2\times P\left(z>1.645\right) \\ P=2\times [1-P\left(z>1.645\right)] \end{gathered}$$

$$\begin{gathered} =2\times [1-0.95] \\ =2\times 0.05 \\ =0.1 \end{gathered}$$

As a result, the value of $p=0.10$, indicating that the sample proportion probability is not equal to $0.467$, but rather $0.467$is $0.10$. As a result, the normal distribution curve depicts the same.

The null hypothesis is not rejected since the alpha value is $0.05$and the p-value is bigger than the alpha value. We don't have enough evidence to say that the proportion of girls born in China isn't $46.7\%$because the null hypothesis isn't rejected.

Let's calculate the $95\%$confidence interval now:

=$p^{\wedge }\pm z$

$$\begin{gathered} =0.4\pm 0.0784 \\ =(0.3216,0.4784) \end{gathered}$$

As a result, the $95\%$confidence level shows that the proportion of Chinese girls born falls between $32.16\%$and $47.84\%$. Let's use the normal distribution curve to depict the same.