Question

Assume, for the sake of this question, that the data were collected through a well-designed, well-implemented random sampling method. The marketing department of a widget manufacturer collected potential consumer preference data regarding a proposed widget upgrade. Three hundred thirty-eight of 575 respondents reported preferring the proposed new widget. The widget manufacturing company had established a threshold of 60% preferring the proposed new widget to move forward with producing the new widgets. At a .10 level of significance, was the threshold probably met

Answer 1

Answer:

Since the pvalue of the test is 0.2743 > 0.1, the threshold probably was met.

Step-by-step explanation:

The widget manufacturing company had established a threshold of 60% preferring the proposed new widget to move forward with producing the new widgets.

This means that at the null hypothesis we test if the proportion is at least 60%, that is:

$H_{0}: p \geq 0.6$

And the alternate hypothesis is:

$H_{a}: p < 0.6$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.6 is tested at the null hypothesis:

This means that:

$\mu = 0.6$

$\sigma = \sqrt{0.6*0.4}$

Three hundred thirty-eight of 575 respondents reported preferring the proposed new widget.

This means that $n = 575, X = \frac{338}{575} = 0.5878$

Value of the test-statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.5878 - 0.6}{\frac{\sqrt{0.6*0.4}}{\sqrt{575}}}$

$z = -0.6$

Pvalue of the test and decision:

We want to find the probability of a proportion of 0.5878 or lower, which is the pvalue of z = -0.6.

Looking at the z-table, z = -0.6 has a pvalue of 0.2743.

Since 0.2743 > 0.1, the threshold probably was met.