Respuesta :
Answer:
Yes, at the 0.10 level of significance, there is enough interest in the service to move to the next planning stage.
Step-by-step explanation:
We are given that the board of directors of a corporation has agreed to allow the human resources manager to move to the next step in planning daycare service for employees' children if the manager can prove that at least 25% of the employees have interest in using the service.
The HR manager polls 300 employees and 90 say they would seriously consider utilizing the service.
Let p = population proportion of employees who seriously consider utilizing the service
SO, Null Hypothesis, [tex]H_0[/tex] : [tex]p \geq[/tex] 25% {means that the service will move to the next planning stage}
Alternate Hypothesis, [tex]H_a[/tex] : p < 25% {means that the service will not move to the next planning stage}
The test statistics that will be used here is One-sample z proportion test statistics because we don't know about the population standard deviation;
T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of employees who seriously consider
utilizing the service = [tex]\frac{90}{300}[/tex] = 0.30
n = sample of employees = 300
So, test statistics = [tex]\frac{0.30-0.25}{\sqrt{\frac{0.30(1-0.30)}{300} } }[/tex]
= 1.889
Now at 0.10 significance level, the z table gives critical value of -1.2816 for left-tailed test. Since our test statistics is more than the critical value of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.
Therefore, we conclude that there is enough interest in the service to move to the next planning stage.
You can use the hypothesis testing to find out if there is enough interest in the service to move to the next planning stage.
We can conclude from the given data, that there is enough interest(at least 25% people are interested) in the service to move to the next planning stage.
How to form the hypotheses?
There are two hypotheses. First one is called null hypothesis and it is chosen such that it predicts nullity or no change in a thing. It is usually the hypothesis against which we do the test. The hypothesis which we put against null hypothesis is alternate hypothesis.
Null hypothesis is the one which researchers try to disprove.
Using above methodology, we form two hypothesis as
Null hypothesis = [tex]H_0: \mu < 25\%\\[/tex]
Alternate hypothesis = [tex]H_1: \mu \geq 25\%[/tex] (one sided, thus, one tailed)
Since we are not given with the standard deviation, we will use one sample z proportion test statistic, which is given as:
[tex]Z = \dfrac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}}[/tex]
where we have:
[tex]\hat{p} = \text{observed proportion} = \dfrac{90}{300} = 0.3\\\\p_0 = \text{null hypothesis value} = 25\% = 0.25\\n =\text{ sample size} = 300\\z = \text{test statistic}[/tex]
or
[tex]Z = \dfrac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}}\\\\Z = \dfrac{0.3 - 0.25}{\sqrt{0.25 \times 0.75/300}} = \dfrac{0.05}{0.025} = 2[/tex]
The p value for z score = 2 for one tailed test is calculated(from calculator) as .02275
Since the p value obtained is < the level of significance 0.1,
thus, we reject the null hypothesis and accept the alternate hypothesis that the average number of people having interest in using the service is at least 25%.
Thus,
We can conclude from the given data, that there is enough interest(at least 25% people are interested) in the service to move to the next planning stage.
Learn more about hypothesis testing here:
https://brainly.com/question/16173147
