Question

A random sample of 150 recent donations at a certain blood bank reveals that 82 were typeA 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 appropriatehypotheses using a significance level of .01. Would your conclusion have been different if asignificance level of .05 had been used?

Answer 1

No conclusion have been not different if a significance level of 0.05 had been used also

The calculated Z-value = 3.596 > 1.96 at 5% level of significance

Null hypothesis is rejected

Alternative Hypothesis is Accepted

The actual percentage of type A donations is not differs from 40%,the percentage of the population having type A blood.

Explanation:

Explanation:-

Step(i):-

A random sample of 150 recent donations at a certain blood bank reveals that 82 were type A blood

Given sample size 'n' = 150

Sample proportion
`p = (x)/(n) = (82)/(150) =0.546`

Given Population proportion P = 40% =0.40

level of significance ∝ = 0.01

Step(ii):-

Null Hypothesis:- H₀: p=0.40

Alternative Hypothesis:-H₁:p≠0.40

The test statistic

`Z =\frac{ p^(-) - p}{\sqrt{(p(1-p))/(n) } }`

a) 99% of level of Z-value

`Z_{(\alpha )/(2) } = Z_{(0.01)/(2) } = Z_(0.005) = 2.576`

`Z =\frac{ 0.546 - 0.40}{\sqrt{(0.546(1-0.546))/(150) } }`

Z = 3.596

The calculated Z-value = 3.596 > 2.576 at 1% level of significance

Null hypothesis is rejected

Alternative Hypothesis is Accepted

The actual percentage of type A donations is not differs from 40%,the percentage of the population having type A blood.

b) 95% of level of Z-value

`Z_{(\alpha )/(2) } = Z_{(0.05)/(2) } = Z_(0.025) = 1.96`

`Z =\frac{ 0.546 - 0.40}{\sqrt{(0.546(1-0.546))/(150) } }`

The calculated Z-value = 3.596 > 1.96 at 5% level of significance

Null hypothesis is rejected

Alternative Hypothesis is Accepted

Conclusion:-

The actual percentage of type A donations is not differs from 40%,the percentage of the population having type A blood.