Answer to Question #241090 in Statistics and Probability for Madiha

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 29 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 = 0.8

Answer 1

We are given that n=11 and X = 29

Therefore, the sample proportion p̂ = 11/29 = 0.38 and population proportion p = 0.7

a)

Here, we are looking for α =P(p>0.7, when p=0.7)

H₀: p=0.7

H_a: p>0.7

we compute the test statistic t as:

α =P(t> (0.7-p)/ $\sqrt{0.7(1-0.7))/11}$ 

=P(t> (0.7 - 0.7)/ $\sqrt{0.7(1-0.7))/11}$ )

= P(t>0)

=1-P(t<0)

For t=0 and df=28, P(t<0)=0.5 from t reference table

Thus

α=1-P(t<0) = 1-0.5 = 0.5

Therefore,

We fail to reject the null hypothesis at α = 0.5

b)

In this case, we are trying to obtain the value of β,assuming that p=0.8

H₀: p=0.8

H_a: p≤0.8

β=P(t≤ (0.8-p)/ $\sqrt{0.7(1-0.7))/11}$ 

=P(t ≤(0.8 - 0.7)/ $\sqrt{0.7(1-0.7))/11}$ ) = 0.72

= P(t≤ 0.72)

For t=0.72 and df=28, P(t≤ 0.72)=1-0.2387 = 0.7613 from t reference table

Thus

β = 0.7613