Respuesta :
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:
[tex]H_{0}: p \geq 0.6[/tex]
And the alternate hypothesis is:
[tex]H_{a}: p < 0.6[/tex]
The test statistic is:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
In which X is the sample mean, [tex]\mu[/tex] is the value tested at the null hypothesis, [tex]\sigma[/tex] is the standard deviation and n is the size of the sample.
0.6 is tested at the null hypothesis:
This means that:
[tex]\mu = 0.6[/tex]
[tex]\sigma = \sqrt{0.6*0.4}[/tex]
Three hundred thirty-eight of 575 respondents reported preferring the proposed new widget.
This means that [tex]n = 575, X = \frac{338}{575} = 0.5878[/tex]
Value of the test-statistic:
[tex]z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{0.5878 - 0.6}{\frac{\sqrt{0.6*0.4}}{\sqrt{575}}}[/tex]
[tex]z = -0.6[/tex]
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.
