Question

The proportion of baby boys born in the United States has historically been 0.508. You choose an SRS of 50 newborn babies and find that 45% are boys. Do ALL calculations to 5 decimal places before rounding.

Answer 1

Answer:

$z=\frac{0.45 -0.508}{\sqrt{\frac{0.508(1-0.508)}{50}}}=-0.82$  

$p_v =P(z<-0.82)=0.206$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to  FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of newborn boys babies is not significantly lower than 0.508

Step-by-step explanation:

Data given and notation

n=50 represent the random sample taken

$\hat p=0.45$ estimated proportion of newborn boys babies

$p_o=0.508$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95  (asumed)

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

Concepts and formulas to use  

We assume that we want to check if the true proportion is less than 0.508.

Null hypothesis:$p\geq 0.508$  

Alternative hypothesis:$p < 0.508$  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.45 -0.508}{\sqrt{\frac{0.508(1-0.508)}{50}}}=-0.82$  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed is $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

$p_v =P(z<-0.82)=0.206$  

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to  FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of newborn boys babies is not significantly lower than 0.508

Answer 2

It is concluded that the null hypothesis H₀ is not rejected. There is not enough evidence to claim that the population proportion p is less than 0.508, at the α =0.05 significance level.

What is the z test statistic for one sample proportion?

Suppose that we have:

• n = sample size
• $\hat{p}$ = sample proportion
• p₀ = population proportion (hypothesised)

Then, the z test statistic for one sample proportion is:

$Z = \dfrac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1-p_0)}{n}}}$

Making the hypothesis and performing the test, assuming that we want to test if the population mean proportion is < 0.508 at the level of significance 0.05.

• (1) Null and Alternative Hypotheses

The following null and alternative hypotheses for the population proportion needs to be tested:

$H_0: p \geq 0.508 \\\\ H_a: p & < & 0.508 \end{array}$

This corresponds to a left-tailed test, for which a z-test for one population proportion will be used.

• (2) Rejection Region

Based on the information provided, the significance level is $\alpha = 0.05$, and the critical value for a left-tailed test is $z_c = -1.64$

The rejection region for this left-tailed test is $R = \{z: z < -1.645\}$

• (3) Test Statistics

The z-statistic is computed as follows:

$z & = & \displaystyle \frac{\hat p - p_0}{\sqrt{ \displaystyle\frac{p_0(1-p_0)}{n}}} \\\\& = & \displaystyle \frac{0.45 - 0.508}{\sqrt{ \displaystyle\frac{ 0.508(1 - 0.508)}{50}}} \\\\ & = & -0.82$

• (4) Decision about the null hypothesis

Since it is observed that $z = -0.82 \ge z_c = -1.645$,  it is then concluded that the null hypothesis is not rejected.

Thus, it is concluded that the null hypothesis H₀ is not rejected. Therefore, there is not enough evidence to claim that the population proportion p is less than 0.508, at the α =0.05 significance level.

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