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The melting point of each of 16 samples of a certain brand of hydrogenated vegetable oil was determined, resulting in \(\overline x = 94.32\). Assume that the distribution of the melting point is normal with 蟽 =1.20.

a.Test H0: 碌 =95 versus Ha: 碌鈮 95 using a two -tailed level .01 test.

b.If a level .01 test is used, what is 尾(94), the probability of a type II error when 碌=94?

c.What value of n is necessary to ensure that 尾(94) = .1 when 伪 = .01?

Short Answer

Expert verified

a)The P value\({\rm{0}}{\rm{.0232}}\) is bigger than \(\alpha {\rm{ = 0}}{\rm{.01}}\), do not reject hypothesis \({H_0}\).

b) \(\beta (94) = 0.2266\)

c) \(n = 22\)

Step by step solution

01

Step 1:Null hypothesis.

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 Ha, 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.

02

Step 2:Solution for part a).

The z values are taken from the appendix table A.3 as well as \(\Phi (z)\) values.

Testing \({H_0}:\mu = 95\) versus \({H_a}:\mu \ne 95\) based on a sample of size \(n = 16\), which is from a normal population with standard deviation \(\sigma = 1.2\).

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

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

The value of the statistic, for \(\overline x = 94.32\), the test statistic value is,

\(\begin{array}{l}z = \frac{{\overline x - {\mu _0}}}{{\sigma /\sqrt n }}\\z = \frac{{94.32 - 95}}{{1.2/\sqrt {16} }}\\z = - 2.27\end{array}\)

03

Step 3:Solution for part a): P value.

The P value for the two sides alternative hypothesis is,

\(\begin{array}{l}{\rm{P(Z > - z and Z < z) = P(Z > 2}}{\rm{.27 and Z < - 2}}{\rm{.27)}}\\{\rm{ = 2}}\Phi {\rm{( - 2}}{\rm{.27)}}\\{\rm{ = 2(0}}{\rm{.0116)}}\\{\rm{ = 0}}{\rm{.0232}}\end{array}\)

The P value\({\rm{0}}{\rm{.0232}}\) is bigger than \(\alpha {\rm{ = 0}}{\rm{.01}}\), do not reject hypothesis \({H_0}\).

04

Step 4:Solution for part b).

Using the two-tailed level \(0.01\) test, then

\(\begin{array}{l}{z_{\alpha /2}} = {z_{0.005}}\\ = 2.58\end{array}\)

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

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

The type II error when \(\mu ' = 94\) is,

\(\begin{array}{l}\beta (94) = \Phi \left( {2.58 + \frac{{95 - 94}}{{1.2/\sqrt {16} }}} \right) - \Phi \left( { - 2.58 - \frac{{95 - 94}}{{1.2/\sqrt {16} }}} \right)\\ = \Phi (5.91) - \Phi (0.75)\\ = 0.9999 - 0.7733\\\beta (94) = 0.2266\end{array}\)

05

Step 5:Solution for part c).

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 \({z_{\alpha /2}} = 2.58\) and for \(\beta = 0.01,{z_\beta } = 1.28\)

Hence the necessary sample size is

\(\begin{array}{l}n = {\left( {\frac{{1.2(2.58 + 1.28)}}{{95 - 94}}} \right)^2}\\n = 21.46\end{array}\)

But the integer is needed, round it up to \(n = 22\).

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