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The null hypothesis, denoted by H₀, is the claim that is initially assumed to be true (the 鈥減rior belief鈥� claim). The alternative hypothesis, denoted by H_a, is the assertion that is contradictory to H₀.

The null hypothesis will be rejected in favour of the alternative hypothesis only if sample evidence suggests that H₀ is false. If the sample does not strongly contradict H₀, we will continue to believe in the plausibility of the null hypothesis. The two possible conclusions from a hypothesis-testing analysis are then reject H0 or fail to reject H₀.

Testing \({H_0}:\mu = 30,000\) versus \({H_a}:\mu > 30,000\)based on the sample of size \(n = 16\), which is from a normal population with standard deviation \(\sigma = 1500\)

assumption that the sample is from normal population distribution with known standard deviation allows using test statistic,

\(Z = \frac{{\overline X - {\mu _0}}}{{\sigma /\sqrt n }}\)

In this case \({\mu _0} = 30,000\) and \(\overline x = 30,960\), the test statistic value is,

\(\begin{array}{l}z = \frac{{\overline x - {\mu _0}}}{{\sigma /\sqrt n }}\\z = \frac{{30,960 - 30,000}}{{1500/\sqrt {16} }}\\z = 2.56\end{array}\)

The alternative hypothesis is \({H_a}:\mu > 30,000\), therefore the area under the standard normal curve to the left of z is needed in order to obtain the P value. The P value is,

\(\begin{array}{l}P = P(Z \ge z)\\ = P(Z \ge 2.56)\\ = 1 - P(Z < 2.56)\\ = 1 - \Phi (2.56)\end{array}\)

\(P = 0.0052\)

From the appendix of the book (you could use a software to compute the value as well).

Since \(0.0052 < \alpha = 0.01\), reject null hypothesis.

Type II error probability\(\beta (\mu ')\)for a level\(\alpha \)test, when alternative hypothesis is\({H_a}:\mu > {\mu _0}\), is

\(\beta (\mu ') = \Phi \left( {{z_\alpha } + \frac{{{\mu _0} - \mu '}}{{\sigma /\sqrt n }}} \right)\)

The alternative hypothesis is\({H_a}:\mu > 30,000\), therefore type II error, for\({z_\alpha } = 2.33\)(computed by a software or taken from the appendix ) is,

\(\begin{array}{l}\beta (30,500) = \Phi \left( {2.33 + \frac{{30,000 - 30,500}}{{1500/\sqrt {16} }}} \right)\\ = \Phi (1)\\\beta (30,500) = 0.8413\end{array}\)

The required sample size\(n\)for which a level\(\alpha \)test produces\(\beta (\mu ') = \beta \)for upper or lower test is,

\(n = {\left( {\frac{{\sigma ({z_\alpha } + {z_\beta })}}{{{\mu _0} - \mu '}}} \right)^2}\)

From the appendix or a software for\(\alpha = 0.01,{z_\alpha } = 2.33\)and for \(\beta = 0.05,{z_\beta } = 1.645\)

Hence the necessary sample size is

\(\begin{array}{l}n = {\left( {\frac{{\sigma ({z_\alpha } + {z_\beta })}}{{{\mu _0} - \mu '}}} \right)^2}\\n = {\left( {\frac{{1500(2.33 + 1.645)}}{{30,000 - 30,500}}} \right)^2}\\n = 142.2\end{array}\)

But the integer is needed, which means that \(n = 143\).

As calculated in (a), the P value is

\(P = 0.0052\)

And the null hypothesis is rejected for \(0.0052 \le \alpha \), which indicates that the smallest \(\alpha \) at which \({H_0}\) can be reject is \(\alpha = 0.0052\).