Answer to Question #241448 in Statistics and Probability for Reyad

Question

A dry-cleaning establishment claims that a new spot remover will remove more

than 70% of the spots to which it is applied. To check this claim, the spot remover

will be used on ‘55’ spots chosen at random. If fewer than 11 of the spots are

removed, we shall not reject the null hypothesis that p = 0.7; otherwise, we conclude that p > 0.7.

i. Evaluate α, assuming that p = 0.7.

ii. Evaluate β for the alternative p = 0.9.

Answer 1

Let the random variable x represent the number of spots that are removed by the spot remover.

Let n=12 represents the number of spots that are randomly selected.

Let p=0.80 represent the proportion of spots removed by spot remover.

Hence, the random variable X follows binomial distribution with parameter p=0.80.

The probability mass function of the random variable X is given by,

$P(X=x) = \binom{n}{x}p^x(1-p)^{n-x}$ x = 0,1,2,…,n

$= \binom{12}{x}(0.80)^x(1-0.80)^{12-x}$ x = 0,1,2,…,12

The critical region is x ≥11

(i) Rejection of the null hypothesis when it is true is known as type-I error.

The probability of committing type-I error is known as level of significance which is denoted by α.

$α = P(type \;I \;error) \\ = P(x ≥11 \;when \; p=0.80) \\ = \sum^{12}_{x=11}b(x;12,0.8) \\ = \sum^{12}_{x=11} \binom{12}{x}(0.8)^x(1-0.8)^{12-x} \\ = \binom{12}{11}(0.8)^{11}(1-0.8)^{12-11} + \binom{12}{12}(0.8)^{12}(1-0.7)^{12-12} \\ = 0.20615 + 0.06872 \\ = 0.27487$

(ii) Fail to reject the null hypothesis when it is false is known as type-II error.

The probability of committing type-II error is denoted by β.

$β = P(type \;II \;error) \\ = P(x<11 \;when \;p=0.9) \\ = \sum^{10}_{x=0}b(x;12,0.9) \\ = 1 -[ \sum^{12}_{x=11} \binom{12}{x} (0.9)^x(1-0.9)^{12-x} ] \\ = 1 -[ \binom{12}{11} (0.9)^{11}(1-0.9)^{12-11} + \binom{12}{12} (0.9)^{12}(1-0.9)^{12-12}] \\ = 1 -[ 0.3766 + 0.2824 ] \\ = 1 -0.6590 \\ = 0.3410$