Respuesta :
Answer:
There is not enough evidence to support the executive's claim that the percentage is actually different from the reported percentage of 26%.
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 100
p = 26% = 0.26
Alpha, α = 0.01
First, we design the null and the alternate hypothesis
[tex]H_{0}: p = 0.26\\H_A: p \neq 0.26[/tex]
This is a two-tailed test.
Formula:
[tex]\hat{p} = 0.17[/tex]
[tex]z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
Putting the values, we get,
[tex]z = \displaystyle\frac{0.17-0.26}{\sqrt{\frac{0.26(1-0.26)}{100}}} = -2.051[/tex]
Now, we calculate the p-value from the table.
P-value = 0.040267
Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.
Thus, there is not enough evidence to support the executive's claim that the percentage is actually different from the reported percentage of 26%.
