Chapter 8.1, Problem 11E

a.

To determine

Explain the Type I error and Type II errors in the context of this problem situation.

Explanation of Solution

Given info:

In a certain region two different companies have applied to provide cable television service. Suppose p be the proportion of all potential subscribers who favor the first company over the second. The null hypothesis is $H_0:p=0.5$ versus the alternative hypothesis $H_a:p\ne 0.5$. The sample size is 25. Let X be the number in the sample who favor the first company.

Justification:

Type I error:

Reject the null hypothesis when it is actually true.

Type II error:

Fail to reject the null hypothesis when it is actually false.

Hypotheses:

Null hypothesis:

$H_0:p=0.5$

Alternative hypothesis:

$H_a:p\ne 0.5$

If the hypothesis concluded that the proportion of all potential subscribers who favor the first company over the second differs from 50-50 proportion, but in reality, the proportion of all potential subscribers who favor the first company over the second is 50-50, then Type I error will arise.

If the hypothesis concluded that the proportion of all potential subscribers who favor the first company over the second is 50-50 proportion, but in reality, the proportion of all potential subscribers who favor the first company over the second differs from 50-50, then Type II error will arise.

b.

To determine

Find the values of X which will be at least as contradictory to $H_0$ as for x = 6.

Answer to Problem 11E

The all possible values which are at least contradictory as x = 6 are $\left\{0,1,2,3,4,5,6,19,20,21,22,23,24,25\right\}$

Explanation of Solution

Calculation:

Assume X be the number in the sample who favor the first company. The samples are randomly selected with two possible outcomes.

Hence, X follows binomial distribution, $X\sim Bin\left(n=25,p=0.5\right)$ .

It is known that the mean of binomial random variable is $E\left(X\right)=np$.

If the null hypothesis $H_0:p=0.5$ is true, then the expected success will be

$\begin{array}{c}np=25\times 0.5\\ =12.5\end{array}$

Here, the values which are less than 6 are also as contradictory to reject the null hypothesis $H_0$.

The probability mass function of binomial distribution for n number of trials and for success probability p is,

$P\left(X=x\right)=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x},x=0,\mathrm{1..}n$

The distance of 6 from 12.5 is $12.5-6=6.5$ the same distance to the right side is $12.5+6.5=19$ .

Hence, it can be said that the right side are also will be equally contradictory for 19 or more values.

Hence, the all possible values which are at least contradictory as x = 6 are $\left\{0,1,2,3,4,5,6,19,20,21,22,23,24,25\right\}$

c.

To determine

Find the probability distribution of the test statistic when $H_0$ is true.

Find the P-value for x = 6.

Answer to Problem 11E

The distribution of X is,

$P\left(X=x\right)=\left(\begin{array}{l}25\\ x\end{array}\right){\left(0.5\right)}^x{\left(1-0.5\right)}^{25-x},x=0,\mathrm{1..},25$.

The P-value for x = 6 is 0.014.

Explanation of Solution

Calculation:

From part b, $X\sim Bin\left(n=25,p=0.5\right)$ and the null hypothesis will be rejected if the values are $\left\{0,1,2,3,4,5,6,19,20,21,22,23,24,25\right\}$.

The null hypothesis is defined as $H_0:p=0.5$.

Thus, when the null hypothesis is true, then $p=0.5$.

Hence, when $H_0$ is true is $X\sim Bin\left(n=25,p=0.5\right)$.

That is,

$P\left(X=x\right)=\left(\begin{array}{l}25\\ x\end{array}\right){\left(0.5\right)}^x{\left(1-0.5\right)}^{25-x},x=0,\mathrm{1..},25$

P-value:

The probability of getting the value of the statistic that is as extreme as the observed statistic when the null hypothesis is true is called as p-value. Therefore, it assumes “null hypothesis is true”.

The P-value is,

$\begin{array}{c}P\left(X\le 6\text{or}X\ge 19\right)=P\left(X\le 6\right)+P\left(X\ge 19\right)\\ =P\left(X\le 6\right)+1-P\left(X\le 18\right)\\ =B\left(6;25,0.5\right)+1-B\left(18;25,0.5\right)\end{array}$

Where,$B\left(x;n,p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Procedure for binomial distribution table value:

From the table A.1 of Cumulative Binomial probabilities,

• Locate n = 25
• Along with n = 25, choose x = 6, 18
• Then, obtain the table value corresponding to p = 0.5.

The value of $B\left(6;25,0.5\right)$ is 0.007and $B\left(18;25,0.5\right)$ is 0.993.

Hence,

$\begin{array}{c}P\left(X\le 6orX\ge 19\right)=B\left(6;25,0.5\right)+1-B\left(18;25,0.5\right)\\ =0.007+1-.993\\ =0.014\end{array}$

The P-value for x = 6 is 0.014.

d.

To determine

Find the probability of Type II error when p = 0.4, 0.3, 0.6 and 0.7.

Answer to Problem 11E

The probability of Type II error for = 0.4, 0.3, 0.6, 0.7 are 0.846, 0.488, 0.846 and 0.488 respectively.

Explanation of Solution

Given info:

$H_0$ will be rejected when $P-\text{value}\le \text{0}\text{.044}$.

Calculation:

The P-value for two-tailed test is divided into two sides of the hypothesis test.

Thus, for two tailed test $P-\text{value}\le \text{0}\text{.044}$ means that in table $P-\text{value}\le \text{2}\times \text{0}\text{.022}$.

Procedure for binomial distribution table value:

From the table A.1 of Cumulative Binomial probabilities,

• Locate n = 25
• Along with n = 25, choose p = 0.5.
• Locate the probability 0.022.
• The value of x is 7.

The distance of 7 from 12.5 is $12.5-7=5.5$ the same distance to the right side is $12.5+5.5=18$

Hence, $H_0$ will be rejected when $P-\text{value}\le \text{0}\text{.044}$ similar to $H_0$ will be rejected when $X\le 7orX\ge 18$

Type II error:

Fail to reject the null hypothesis when it is actually false.

The rejection rule of null hypothesis is $X\le 7orX\ge 18$.

Hence for fail to reject the null hypothesis the range of x will be $7<X<18$.

Type II error for p:

$\begin{array}{c}\beta \left(p\right)=P\left(\text{Fail to reject the null hypothesis}\right)\\ =P\left(7<X<18\right)\\ =P\left(X\le 17\right)-P\left(X\le 7\right)\\ =B\left(17;25,p\right)-B\left(7;25,p\right)\end{array}$

Where,$B\left(x;n,p\right)={\displaystyle \sum _{y=0}^{x}b\left(y;n,p\right)}$

Type II error for p = 0.4:

Procedure for binomial distribution table value:

From the table A.1 of Cumulative Binomial probabilities,

• Locate n = 25
• Along with n = 25, choose x = 17, 7
• Then, obtain the table value corresponding to p = 0.4.

The value of $B\left(17;25,0.4\right)=0.999$ $B\left(7;25,0.4\right)=0.154$

Hence,

$\begin{array}{c}\beta \left(0.4\right)=B\left(17;25,0.4\right)-B\left(7;25,0.4\right)\\ =0.999-0.154\\ =0.846\end{array}$

Type II error for p = 0.3:

Procedure for binomial distribution table value:

From the table A.1 of Cumulative Binomial probabilities,

• Locate n = 25
• Along with n = 25, choose x = 17, 7
• Then, obtain the table value corresponding to p = 0.3.

The value of $B\left(17;25,0.3\right)=1$ $B\left(7;25,0.3\right)=0.512$

Hence,

$\begin{array}{c}\beta \left(0.3\right)=B\left(17;25,0.3\right)-B\left(7;25,0.3\right)\\ =1-0.512\\ =0.488\end{array}$

Type II error for p = 0.6:

Procedure for binomial distribution table value:

From the table A.1 of Cumulative Binomial probabilities,

• Locate n = 25
• Along with n = 25, choose x = 17, 7
• Then, obtain the table value corresponding to p = 0.6.

The value of $B\left(17;25,0.6\right)=0.846$ $B\left(7;25,0.6\right)=0$

Hence,

$\begin{array}{c}\beta \left(0.6\right)=B\left(17;25,0.6\right)-B\left(7;25,0.6\right)\\ =0.846-0\\ =0.846\end{array}$

Type II error for p = 0.7:

Procedure for binomial distribution table value:

From the table A.1 of Cumulative Binomial probabilities,

• Locate n = 25
• Along with n = 25, choose x = 17, 7
• Then, obtain the table value corresponding to p = 0.7.

The value of $B\left(17;25,0.7\right)=0.488$ $B\left(7;25,0.7\right)=0$

Hence,

$\begin{array}{c}\beta \left(0.7\right)=B\left(17;25,0.7\right)-B\left(7;25,0.7\right)\\ =0.488-0\\ =0.488\end{array}$

Hence, the probability of Type II error for = 0.4, 0.3, 0.6, 0.7 are 0.846, 0.488, 0.846 and 0.488 respectively.

e.

To determine

Find the conclusion if 6 of the 25 queried favored company 1.

Answer to Problem 11E

The proportion of all potential subscribers who favor the first company over the second is not 50-50 proportion.

Explanation of Solution

Calculation:

If 6 of the 25 queried favored company 1, then the P-value is $P\left(X\le 6orX\ge 19\right)$

From part c, $P\left(X\le 6orX\ge 19\right)=0.014$

Decision rule:

If $P-\text{value}\le \text{0}\text{.044}$ , reject the null hypothesis $H_0$

If $P-\text{value > 0}\text{.044}$ , fail to reject the null hypothesis $H_0$

Conclusion:

Here, the P-value is less than 0.044

That is, $P\text{-value}\left(=0.014\right)<0.044$.

By rejection rule, “reject null hypothesis”.

Hence, the null hypothesis will be rejected.

Hence, the proportion of all potential subscribers who favor the first company over the second is not 50-50 proportion.