Question

A random sample of 147 recent donations at a certain blood bank reveals that 87 were type A blood. Does this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood? Carry out a test of the appropriate hypotheses using a significance level of 0.01. State the appropriate null and alternative hypotheses.

Answer 1

Answer:

$z=\frac{0.592 -0.4}{\sqrt{\frac{0.4(1-0.4)}{147}}}=4.75$  

$p_v =2*P(z>4.75)=0.000002034$  

The p value is lower than the significance level of 0.01. So then we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of type A donations differs from 40%, the percentage of the population having type A blood

Step-by-step explanation:

Infomration given

n=147 represent the random sample taken

X=87 represent the people who are type A blood

$\hat p=\frac{87}{147}=0.592$ estimated proportion of people with type A blood

$p_o=0.4$ is the value to verify

$\alpha=0.01$ represent the significance level

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to verify if type A donations differs from 40% so then the system of hypothesis are:  

Null hypothesis:$p=0.4$  

Alternative hypothesis:$p \neq 0.4$  

The statistic for this case is given:

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

Replacing the info given we got:

$z=\frac{0.592 -0.4}{\sqrt{\frac{0.4(1-0.4)}{147}}}=4.75$  

Now we can calculate the p value with this probability:

$p_v =2*P(z>4.75)=0.000002034$  

The p value is lower than the significance level of 0.01. So then we have enough evidence to reject the null hypothesis and we can conclude that the true proportion of type A donations differs from 40%, the percentage of the population having type A blood

Answer 2

Answer:

Step-by-step explanation:

Since we have given n = 157

x = 86

So,

$\hat{p}=\dfrac{x}{n}=\dfrac{87}{147}=0.59$

and we have p = 0.4

So, hypothesis would be

$H_0:p=\hat{p}\\\\H_a:p\neq \hat{p}$

Since there is 1% level of significance.

So, test statistic value would be

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.59-0.40}{\sqrt{\dfrac{0.4\times 0.6}{147}}}\\\\z=\dfrac{0.19}{0.0404}\\\\z=4.70$

Now we can calculate the p value with this probability:

 =2*P(z>4.70)=0.00000203

The p value is lower than the significance level of 0.01

So, we reject the null hypothesis.

Hence, Yes, this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood.