Chapter 3, Problem 102SE

An educational consulting firm is trying to decide whether high school students who have never before used a hand-held calculator can solve a certain type of problem more easily with a calculator that uses reverse Polish logic or one that does not use this logic. A sample of 25 students is selected and allowed to practice on both calculators. Then each student is asked to work one problem on the reverse Polish calculator and a similar problem on the other. Let p = P(S), where S indicates that a student worked the problem more quickly using reverse Polish logic than without, and let X = number of S's.

1. a. If p = .5, what is P(7 ≤ X ≤ 18)?
2. b. If p = .8. what is P(7 ≤ X ≤ 18)?
3. c. If the claim that p = .5 is to be rejected when either x ≤ 7 or x ≥ 18, what is the probability of rejecting the claim when it is actually correct?
4. d. If the decision to reject the claim p = .5 is made as in part (c), what is the probability that the claim is not rejected when p = .6? When p = .8?
5. e. What decision rule would you choose for rejecting the claim p = .5 if you wanted the probability in part (c) to be at most .01?

To determine

Find the value of $P\left(7\le X\le 18\right)$ for $p=0.5$

Answer to Problem 102SE

The value of $P\left(7\le X\le 18\right)$ is 0.986.

Explanation of Solution

Given info:

An educational consulting firm wants to check whether the high school students who have never used a hand held calculator can solve a certain type of problem more easily with a calculator that uses reverse polish logic or one that does not use the logic. 25 students were selected as a sample. The both type of calculators was used by 25 students. $p=P\left(S\right)$, where S indicates the student worked the problem more quickly using reverse Polish logic than without. X be the number of S’s

Calculation:

The value of $P\left(7\le X\le 18\right)$:

Let X be the number of S’s. p = 0.5.

Hence, $X\sim Bin\left(25,0.5\right)$

It is known that $P\left(a\le X\le b\right)=F\left(b\right)-F\left(a-1\right)$

Using this formula

$\begin{array}{c}P\left(7\le X\le 18\right)=P\left(X\le 18\right)-P\left(X\le 6\right)\\ =B\left(18;25,0.5\right)-B\left(6;25,0.5\right)\end{array}$

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 = 18, 6
• Then obtain the table value corresponding to p = 0.5

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

Hence,

$\begin{array}{c}P\left(7\le X\le 18\right)=B\left(18;25,0.5\right)-B\left(6;25,0.5\right)\\ =0.993-0.007\\ =0.986\end{array}$

Thus, the value of $P\left(7\le X\le 18\right)$ is 0.986.

To determine

Find the value of $P\left(7\le X\le 18\right)$ for $p=0.8$

Answer to Problem 102SE

The value of $P\left(7\le X\le 18\right)$ is 0.220.

Explanation of Solution

Calculation:

The value of$P\left(7\le X\le 18\right)$:

Let X be the number of S’s. p = 0.8.

Hence, $X\sim Bin\left(25,0.8\right)$

$\begin{array}{c}P\left(7\le X\le 18\right)=P\left(X\le 18\right)-P\left(X\le 6\right)\\ =B\left(18;25,0.8\right)-B\left(6;25,0.8\right)\end{array}$

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 = 18, 6
• Then, obtain the table value corresponding to p = 0.8

The value of $B\left(18;25,0.8\right)=0.220,B\left(6;25,0.8\right)=0$= 0.993

Hence,

$\begin{array}{c}P\left(7\le X\le 18\right)=B\left(18;25,0.8\right)-B\left(6;25,0.8\right)\\ =0.220-0\\ =0.220\end{array}$

Thus, the value of $P\left(7\le X\le 18\right)$ is 0.220.

To determine

Find the probability of rejecting the claim when it is actually correct.

Answer to Problem 102SE

The probability of rejecting the claim when it is actually correct is 0.044.

Explanation of Solution

Given info:

The claim is that p = 0.5 is to be rejected when either $x\le 7orx\ge 18$

Calculation:

The probability of rejecting the claim when it is actually correct:

$\begin{array}{c}P\left(X\le 7\right)+P\left(X\ge 18\right)=P\left(X\le 7\right)+1-P\left(X\le 17\right)\\ =B\left(7;25,0.5\right)+1-B\left(17;25,0.5\right)\end{array}$

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 = 7, 17
• Then, obtain the table value corresponding to p = 0.5

The value of $B\left(7:25,0.5\right)=0.022$ and $B\left(17;25,0.5\right)=0.978$

Hence,

$\begin{array}{c}P\left(X\le 7\right)+P\left(X\ge 18\right)=0.022+1-0.978\\ =0.022+0.022\\ =0.044\end{array}$

Thus, the probability of rejecting the claim when it is actually correct is 0.044.

To determine

Find the probability that the claim in part c is not rejected for p = 0.6 and p = 0.8

Answer to Problem 102SE

The probability that the claim in part c is not rejected for p = 0.6 is 0.845.

The probability that the claim in part c is not rejected for p = 0.8 is 0.109.

Explanation of Solution

Calculation:

Here, the values of p = 0.6 and p = 0.8

The claim is that p = 0.6 is to be rejected when either $x\le 7orx\ge 18$

The rule for non-rejection is $7<x<18$

The probability of not rejecting the claim when p = 0.6:

$\begin{array}{c}P\left(7<X<18\right)=P\left(8\le X\le 17\right)\\ =P\left(X\le 17\right)-P\left(X\le 7\right)\\ =B\left(17;25,0.6\right)-B\left(7;25,0.6\right)\end{array}$

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 = 7, 17
• Then, obtain the table value corresponding to p = 0.6

The value of $B\left(7:25,0.6\right)=0.001$ and $B\left(17;25,0.6\right)=0.846$

Hence,

$\begin{array}{c}P\left(8\le X\le 17\right)=B\left(17;25,0.6\right)-B\left(7;25,0.6\right)\\ =0.846-0.001\\ =0.845\end{array}$

Thus, the probability that the claim in part c is not rejected for p = 0.6 is 0.845.

The probability of not rejecting the claim when p = 0.8:

$\begin{array}{c}P\left(7<X<18\right)=P\left(8\le X\le 17\right)\\ =P\left(X\le 17\right)-P\left(X\le 7\right)\\ =B\left(17;25,0.8\right)-B\left(7;25,0.8\right)\end{array}$

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 = 7, 17
• Then, obtain the table value corresponding to p = 0.8

The value of $B\left(7:25,0.8\right)=0$ and $B\left(17;25,0.8\right)=0.109$

Hence,

$\begin{array}{c}P\left(8\le X\le 17\right)=B\left(17;25,0.8\right)-B\left(7;25,0.8\right)\\ =0.109-0\\ =0.109\end{array}$

Thus, the probability that the claim in part c is not rejected for p = 0.8 is 0.109.

To determine

Find the decision rule for rejecting the claim if the probability in part c is at most 0.01.

Answer to Problem 102SE

The decision rule for rejecting the claim is $x\le 5orx\ge 20$ if probability in part c is at most 0.01.

Explanation of Solution

Calculation:

Here, the probability in part c, is at most 0.01.

The condition is $P\left(\text{rejecting the claim when }p\text{=0}\text{.5}\right)\le 0.01$

For the range, $x\le 7orx\ge 18$, the probability is 0.004 (from part c).

For the range,$x\le 6orx\ge 19$

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

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 $B\left(6:25,0.5\right)=0.007$ and $B\left(18;25,0.5\right)=0.993$

Hence,

$\begin{array}{c}P\left(X\le 6\right)+P\left(X\ge 19\right)\\ =B\left(6;25,0.5\right)+1-B\left(18;25,0.5\right)\\ =0.007+1-0.993\\ =0.014\end{array}$

The value 0.014 is too large.

For the range,$x\le 5orx\ge 20$

$\begin{array}{c}P\left(X\le 5\right)+P\left(X\ge 20\right)=P\left(X\le 5\right)+1-P\left(X\le 19\right)\\ =B\left(5;25,0.5\right)+1-B\left(19;25,0.5\right)\end{array}$

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 = 5, 19
• Then, obtain the table value corresponding to p = 0.5

The value of $B\left(5:25,0.5\right)=0.002$ and $B\left(19;25,0.5\right)=0.998$

Hence,

$\begin{array}{c}P\left(X\le 5\right)+P\left(X\ge 20\right)=B\left(5;25,0.5\right)+1-B\left(19;25,0.5\right)\\ =0.002+1-0.998\\ =0.004\end{array}$

The value 0.004 is less than 0.01.

Hence, the decision rule for rejecting the claim if the probability in part c is at most 0.01 is $x\le 5orx\ge 20$.