Answer to Question #155072 in Statistics and Probability for Ruth

Question

Question 4. 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

The population proportion

$p_0= 0.3$

The sample proportion

$\hat p =\frac{16}{60}=0.2667$

Part a)

Null hypothesis, $H_0: \hat p = 0.3000$

Alternative hypothesis, $H_1: \hat p \not=0.3000$

Part b)

Type I error will occur when we conclude that the proportion of customers that were able to program their DVD is not equal to $30\%$ yet the true proportion is $30\%$

Type II occurs when we conclude that the proportion of customers that were able to program their DVD is equal to $30\%$ when the true proportion is actually not equal to $30\%$

Part c)

$$Z=\frac{\hat p-p_0}{ \sqrt{\frac{p_0(1-p_0)}{n}}}$$

$$Z=\frac{0.2667-0.3}{ \sqrt{\frac{0.3(1-0.3)}{60}}}= -0.5629$$

The p value corresponding to the Z value obtained is

$$p=0.287$$

This is a two tail test. At a $0.05$ level of significance, we fail to reject the null hypothesis if the p value obtained is in the range

$$0.025<p\ value<0.975$$

Since our p value = 0.287 is in the acceptance region, we fail to reject the null hypothesis and conclude that the proportion of customers that were able to program their DVD players was not affected by the new booklet.

The p-value is obtained from the normal distribution tables. The p-value corresponding to the Z score calculated from the data is identified from the normal distribution table. The z value is obtained as the number of standard deviations of the sample mean, away from the true mean when standardized to a distribution with mean = 0 and standard deviation = 1.

The range of p-values is obtained by considering the standard normal tables and the required level of significance. At a 5% level of confidence, we expect to reject the most extreme 5% of sample means. Since we are testing whether the sample and population means are the same, we use a two-tail test to obtain the range of p-values. In our case, the 5% that is rejected is divided equally between both the lowest 2.5% and the highest 2.5% of sample means. These are the values corresponding to a cummlative probability of between 0.025 and 0.975