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A random sample of \(150\) recent donations at a certain blood bank reveals that \(82\) 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 \(.01\). Would your conclusion have been different if a significance level of \(.05\) had been used?

Short Answer

Expert verified

The null hypothesis has been rejected.

Step by step solution

01

Define p-value in hypothesis testing.

The null hypothesis states that the population mean is equal to the value mentioned in the claim. If the null hypothesis is the claim, then the alternative hypothesis states the opposite of the null hypothesis.

\(\begin{array}{l}{H_0}:p = 0\\{H_a}:p \ne 0\end{array}\)

The formula for the value of the test statistic is given by,\(z = \frac{{\hat p - {p_0}}}{{\sqrt {\frac{{{p_0}\left( {1 - {p_0}} \right)}}{n}} }}\).

The sample proportion is calculated by dividing the number of successes by the sample size. \(\hat p = \frac{x}{n}\)

02

Test the appropriate hypothesis.

Let the appropriate hypotheses be\({H_0}:p = 0.4\)versus\({H_0}:p \ne 0.4\).

Let the given be:

\(\begin{aligned}{c}{p_0} &= 0.4\\n &= 150\end{aligned}\)

Sample proportion:

\(\hat p = \frac{{82}}{{150}}\)

The value of the test-statistic:

\(\begin{aligned}{c}z &= \frac{{82/150 - 0.4}}{{\sqrt {0.4(1 - 0.4)/150} }}\\ &= \frac{{0.147}}{{0.04}}\\ &= 3.667\end{aligned}\)

The P-value of the alternative hypothesis\({H_a}:p \ne {p_0}\)is,

\(\begin{array}{l} = 2 \times (Area under the standard normal curve to the right of|z|)\\ = 2 \times P(z \ge 3.667)\\ = 2 \times 0\\ = 0\end{array}\)

Hence, the null hypothesis has been rejected because \(0 < 0.01\). The percentage of all donors with blood type A differs from \(40\% \). The same conclusions would be the same for \(\alpha = 0.05\).

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