Question

A random sample of 150recent donations at a certain

blood bank reveals that
45 were type A blood.Does this suggest that the
actual percentage of type A
donations is less than40%, the percentage of the
population having type A
blood? Carry out a testof the appropriate
hypotheses using a significance
level of0.01.

Answer 1

Null hypothesis:
`p\geq 0.4`

Alternative hypothesis:
`p < 0.4`

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

`p_v =2*P(Z<-2.5)=0.0124`

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.

Explanation:

1) Data given and notation

n=150 represent the random sample taken

X=45 represent the people with type A blood

`\hat p=(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{(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{(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<-2.5)=0.0124`

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.