Question

An importer of electronic goods is considering packaging a new, easy-to-read instruction booklet with DVD players. It wants to package this booklet only if it helps customers more than the current booklet. Previous tests found that only 30% of customers were able to program their DVD player. An experiment with the new booklet found that 16 out of 75 customers were able to program their DVD player.

(a) State the null and alternative hypotheses.
(b) Describe Type I and Type II errors in this context.
(c) Find the p-value of the test. Do the data supply enough evidence to reject the null hypothesis if α = 0.05?

Answer 1

H0: p1= p2 against the claim Ha: p1≠p2

The data does not supply enough evidence to reject the null hypothesis .

Explanation:

The null and alternate hypothesis will be

1) H0: p1= p2 against the claim Ha: p1≠p2

Part b)

Type I Error is when H0 is rejected when it is true.

Type II Error is when H0 is accepted when H0 is false.

In this context Type I Error would be when we would reject the H0 for the above calculations when z does not lie in the critical region.

In this context Type II Error would be when we would accept the H0 when z lies in the critical region.

Part C:

2) The significance level is chosen to be ∝=0.05

3) The critical region for two tailed test is z ≥ ± 1.96

4) The test statistic

Z= p1-p2/ sqrt [pcqc( 1/n1+ 1/n2)]

Here p1= 0.3 and p2= 16/75 =0.213

pc = 30+16/100+75

=0.2629

qc= 1-pc= 1-0.2629= 0.7371

5) Calculations

Z= p1-p2/ sqrt [pcqc( 1/n1+ 1/n2)]

z= 0.3-0.213/√ 0.2629*0.7371( 1/100+ 1/30)

z= 1.294

6) Conclusion

Since the calculated value of z= 1.294 does not lie in the critical region z= ± 1.96 the null hypothesis is accepted and it is concluded that there is no difference between the two proportions.

The data does not supply enough evidence to reject the null hypothesis .

The p - value is 0.1957, which is greater than 0.05 it is concluded that the null hypothesis is not rejected.