Question

A random sample of 150recent donations at a certain

Answer 1

Answer:

Null hypothesis:$p\geq 0.4$  

Alternative hypothesis:$p < 0.4$  

$z=\frac{0.3 -0.4}{\sqrt{\frac{0.4(1-0.4)}{150}}}=-2.5$  

$p_v =2*P(Z$  

If we compare the p value obtained and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that the true proportion is not significantly lower than 0.4 or 40% at 1% of significance.  

Step-by-step explanation:

1) Data given and notation  

n=150 represent the random sample taken

X=45 represent the people with type A blood

$\hat p=\frac{45}{150}=0.3$ estimated proportion of people with type A blood

$p_o=0.4$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

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

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion of people type A blood is less than 0.4:  

Null hypothesis:$p\geq 0.4$  

Alternative hypothesis:$p < 0.4$  

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 $p_o$.

3) Calculate the statistic  

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

$z=\frac{0.3 -0.4}{\sqrt{\frac{0.4(1-0.4)}{150}}}=-2.5$  

4) 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 provided $\alpha=0.01$. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =2*P(Z$  

If we compare the p value obtained and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that the true proportion is not significantly lower than 0.4 or 40% at 1% of significance.