Answer to Question #155483 in Statistics and Probability for EUGINE HAWEZA

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 60 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 alpha = 0.05?

Answer 1

(a) The null and alternative hypothesis are:

H₀: p ≤ 0.30

H₁: p > 0.30

(b) A Type I error in this context is that when one rejects null hypothesis which is the proportion of all customers who were able to program their DVD player is 30 % when it is true.

A Type II error in this context is that when one rejects alternative hypothesis which is the proportion of all customers who were able to program their DVD player is more than 30 % when it is true.

(c) Analyze that the data supply does enough evidence to reject the null hypothesis at 5% level of significance.

First, write null and alternative hypothesis, test statistics and p-value then take decision basis of outcome.

The null and alternative hypothesis is given in part (a).

Find the sample statistic \bar{p} and convert it to a z value.

$\bar{p} = \frac{r}{n} \\ = \frac{16}{60} \\ = 0.27$

Let the population proportion be p = 0.30

$q = 1 -p \\ = 1 -0.30 \\ = 0.70$

The formula for the z-test statistic is

$z = \frac{\bar{p}-p}{\sqrt{\frac{pq}{n}}}$

$\bar{p}$ = the sample proportion

p = the population proportion

n = the sample size

$n = 60 \\ z = \frac{0.27-0.30}{\sqrt{\frac{0.30 \times 0.70}{60}}} \\ = -0.50$

The z-test statistics is -0.50

Find the P value for this test by Excel function:

P(z <-0.50) = [=NORMSDIST(-0.50)]

= 0.3085

So, the P value for this test is 0.3085.

It can be conclude that the P value in this context is the p value is greater than the level of significance which is 0.05. So, fail to reject the null hypothesis at 5 % level of significance. There is an insufficient evidence to indicate that the proportion of all customers who were able to program their DVD player is more than 30 %.