Question

Persons having Reynaud’s syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experiment to study the extent of this impairment, each subject immersed a forefinger in water and the resulting heat output \((cal/c{m^2}/min)\) was measured. For \(m = 10\) subjects with the syndrome, the average heat output was \(\bar x = .64\), and for \(n = 10\) non-sufferers, the average output was \(2.05\). Let \({\mu _1}\) and \({\mu _2}\) denote the true average heat outputs for the two types of subjects. Assume that the two distributions of heat output are normal with \({\sigma _1} = .2\) and \({\sigma _2} = .4\).

a. Consider testing \({H_0}:{\mu _1} - {\mu _2} = - 1.0\) versus \({H_2}:{\mu _1} - {\mu _2} < - 1.0\)at level . \(01\). Describe in words what \({H_a}\) says, and then carry out the test.

b. What is the probability of a type II error when the actual difference between \({\mu _1}\) and \({\mu _2}\) is \({\mu _1} - {\mu _2} = - 1.2?\)

c. Assuming that \(m = n\), what sample sizes are required to ensure that \(\beta = .1\) when \({\mu _1} - {\mu _2} = - 1.2?\)

Short answer

the solution is

a) There is enough data to support the claim that the genuine average heat output of subjects with the syndrome is more than 1 lower than that of subjects without the disease.

b) \(\beta = 0.8212 = 82.12\% \)

c) \({\rm{n = 66}}\)

Step-by-step solution

describe \({H_a}\)a)

\(\begin{array}{l}{{\bar x}_1} = 0.64\\{{\bar x}_2} = 2.05\\{\sigma _1} = 0.2\\{\sigma _2} = 0.4\end{array}\)

\(\begin{array}{l}{n_1} = {n_2} = 10\\\alpha = 0.01\\{H_0}:{\mu _1} - {\mu _2} = - 1\end{array}\)

The genuine average heat output of people with the syndrome is one lower than the true average heat output of people who don't have the syndrome.

\({H_a}:{\mu _1} - {\mu _2} < - 1\)

Subjects with the condition have a genuine average heat output that is more than a factor of one lower than subjects without the disease.

Calculate the test statistic's value:

\(z = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{\sigma _1^2}}{{{n_1}}} + \frac{{\sigma _2^2}}{{{n_2}}}} }} = \frac{{(0.64 - 2.05) - ( - 1)}}{{\sqrt {\frac{{0.{2^2}}}{{10}} + \frac{{0.{4^2}}}{{10}}} }} \approx - 2.90\)

If the null hypothesis is true, the P-value is the chance of getting a value that is more extreme or equal to the standardized test statistic $z$. Using the usual probability table, calculate the probability.

\(P = P(Z < - 2.90) = 0.0019\)

The null hypothesis is rejected if the P-value is less than the significance level alpha.

\(P < 0.01 \Rightarrow Reject {H_0}\)

There is enough data to support the claim that the genuine average heat output of subjects with the syndrome is more than 1 lower than that of subjects without the disease.

difference between \({\mu _1}\)and \({\mu _2}\)

b)

\(\begin{array}{l}{{\bar x}_1} = 0.64\\{{\bar x}_2} = 2.05\\{\sigma _1} = 0.2\\{\sigma _2} = 0.4\end{array}\)

\(\begin{array}{l}\alpha = 0.01\\{H_0}:{\mu _1} - {\mu _2} = - 1\\{H_a}:{\mu _1} - {\mu _2} < - 1\\{\Delta _a} = - 1.2\end{array}\)

calculate mean difference

Using the normal probability table in the appendix, find the \(z\)-score that corresponds to a probability of \(\alpha = 0.01\):

\(z = - 2.33\)

The sample mean difference is calculated by multiplying the population mean difference (of the hypothesis) by the product of the \(z\)-score and the standard deviation:

\(\left( {{{\bar x}_1} - {{\bar x}_2}} \right) = \left( {{\mu _1} - {\mu _2}} \right) - {z_{\alpha /2}} \times \sqrt {\frac{{\sigma _1^2}}{{{n_1}}} + \frac{{\sigma _2^2}}{{{n_2}}}} \bar x = - 1 - 2.33\sqrt {\frac{{0.{2^2}}}{{10}} + \frac{{0.{4^2}}}{{10}}} \approx - 1.3295\)

The \(z\)-value is calculated by dividing the sample mean difference by the population mean difference (alternative mean difference! ), then multiplying by the standard deviation:

\(z = \frac{{{{\bar x}_1} - {{\bar x}_2} - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{\sigma _1^2}}{{{n_1}}} + \frac{{\sigma _2^2}}{{{n_2}}}} }} = \frac{{ - 1.3295 - ( - 1.2)}}{{\sqrt {\frac{{0.{2^2}}}{{10}} + \frac{{0.{4^2}}}{{10}}} }} \approx - 0.92\)

When the null hypothesis is false, the probability of making a Type II error is the probability of not rejecting the null hypothesis. Using the normal probability table in the appendix, calculate the chance of rejecting the null hypothesis.

\(\beta = P(Z > - 0.92) = 1 - P(Z < - 0.92) = 1 - 0.1788 = 0.8212 = 82.12\% \)

step 4:calculate mean difference

c)

\(\begin{array}{l}{{\bar x}_1} = 0.64\\{{\bar x}_2} = 2.05\\{\sigma _1} = 0.2\\{\sigma _2} = 0.4\end{array}\)

\(\begin{array}{l}{n_1} = {n_2} = unknown \\\alpha = 0.01\\{H_0}:{\mu _1} - {\mu _2} = - 1\\{H_a}:{\mu _1} - {\mu _2} < - 1\\{\Delta _a} = - 1.2\end{array}\)

\(\beta = 0.10\)

Using the normal probability table in the appendix, find the \({\rm{z}}\)-score that corresponds to a probability of \(\alpha = 0.01\):

\({\rm{z = - 2}}{\rm{.33}}\)

The sample mean difference is calculated by multiplying the population mean difference (of the hypothesis) by the product of the \({\rm{z}}\)-score and the standard deviation:

\(\left( {{{\bar x}_1} - {{\bar x}_2}} \right) = \left( {{\mu _1} - {\mu _2}} \right) + {z_{\alpha /2}} \times \sqrt {\frac{{\sigma _1^2}}{{{n_1}}} + \frac{{\sigma _2^2}}{{{n_2}}}\bar x} = - 1 - 2.33\sqrt {\frac{{0.{2^2}}}{n} + \frac{{0.{4^2}}}{n}} = - 1 - 2.33\frac{{\sqrt {0.{2^2} + 0.{4^2}} }}{{\sqrt n }}\)

Using the normal probability table in the appendix, calculate the \({\rm{z}}\)-score for a probability of \(1 - \beta = 1 - 0.10 = 0.90\):

\(z = 1.28\)

The population mean difference (of the alternative hypothesis) is multiplied by the product of the \({\rm{z}}\)-score and the standard deviation to get the sample mean difference:

\(\left( {{{\bar x}_1} - {{\bar x}_2}} \right) = \left( {{\mu _1} - {\mu _2}} \right) + {z_{\alpha /2}} \times \sqrt {\frac{{\sigma _1^2}}{{{n_1}}} + \frac{{\sigma _2^2}}{{{n_2}}}} \bar x = - 1.2 + 1.28\sqrt {\frac{{0.{2^2}}}{n} + \frac{{0.{4^2}}}{n}} = - 1.2 + 1.28\frac{{\sqrt {0.{2^2} + 0.{4^2}} }}{{\sqrt n }}\)

For the sample mean differences, the two found expressions must be equal:

\( - 1 - 2.33\frac{{\sqrt {0.{2^2} + 0.{4^2}} }}{{\sqrt n }} = - 1.2 + 1.28\frac{{\sqrt {0.{2^2} + 0.{4^2}} }}{{\sqrt n }}\)

step 5:simplify

add \(2.33\frac{{\sqrt {0.{2^2} + 0.{4^2}} }}{{\sqrt n }}\) to each side of the equation:

\( - 1 = - 1.2 + 3.61\frac{{\sqrt {0.{2^2} + 0.{4^2}} }}{{\sqrt n }}\)

add \(1.2\) to each side:

\(0.2 = 3.61\frac{{\sqrt {0.{2^2} + 0.{4^2}} }}{{\sqrt n }}\)

multiply each side by \(\sqrt n \)

\(0.2\sqrt n = 3.61\sqrt {0.{2^2} + 0.{4^2}} \)

divide each side by \(0.2\):

\(\sqrt n = \frac{{3.61\sqrt {0.{2^2} + 0.{4^2}} }}{{0.2}}\)

square each side:

\(n = {\left( {\frac{{3.61\sqrt {0.{2^2} + 0.{4^2}} }}{{0.2}}} \right)^2} \approx 66\)

conclusion

a) There is enough data to support the claim that the genuine average heat output of subjects with the syndrome is more than 1 lower than that of subjects without the disease.

b) \(\beta = 0.8212 = 82.12\% \)

c) \({\rm{n = 66}}\)