Question

3.3 Q: A supplier of digital memory cards claims that a proportion of less than 20% of the cards are defective. In a random sample of 145 memory cards, it is found that 6 are defective. At the 5% level of significance, use the given sample to test the manufacturer’s claim that less than 20% of the items are defective.

Identify the null hypothesis and alternative hypothesis, test statistic, P -value, conclusion about the null hypothesis, and final conclusion that addresses the original claim, by providing the following details:

1. State the hypotheses

1. H₀ : p = 0.20 & H₁: P <0.20; B: p < 0.20 & H₁: P = 0.20; C. H₀: p ≠ 0.20 & H₁: P<0.20;   D. p = 0.20 & H₁: P >0.20

b. Find p˄ (p-hat)

A) 0.04     B) 0.14     C) 0.08     D) 0.17

c. Find he test statistic Z_test(Use Appropriate Calculator Program)

A) 1.124     B) – 4.775     C) - 0.821     D) 0.211       E) -1.700

d. The p-value (Use Calculator Appropriate Calculator Program)

A) 8.99E-7     B) 0.05     C) - 0.821     D) 0.222        E) 0.100

e. The conclusion: Reject H₀ or do not reject H₀ ?

1. Reject H₀;      B. Do not Reject H₀

Answer 1

a) As we are testing here whether less than 20% of the items are defective, therefore the null and the alternative hypothesis here are given as:

b) The point estimate of the sample proportion here is computed here as:

Therefore A) 0.04 is the required point estimate here.

c) The test statistic here is computed as:

Therefore B) - 4.775 is the required test statistic value here.

d) As this is a lower tailed that is a one tailed test, the p-value here is computed from the standard normal tables as:
p = P(Z < -4.7751) = approx. 0

Therefore 0 is the approx. p value here.

e) As the p-value here is approx 0. < 0.05 which is the level of significance, therefore the test is significant here and we Reject the null hypothesis H0 here.