Question

In doing so, you collect a random sample of 50 salespersons employed by his company, which is thought to be representative of sales staffs of competing organizations in the industry. The collected random sample of size 50 showed that only 18 were women. Compute the p-value associated with this test. Place your answer, rounded to 4 decimal places, in the blank. For example, 0.3456 would be a legitimate entry.

Answer 1

Answer:

$z=\frac{0.36 -0.45}{\sqrt{\frac{0.45(1-0.45)}{50}}}=-1.279$  

$p_v =P(z$

And we can use the following code to find it  "=NORM.DIST(-1.279,0,1,TRUE)"

Step-by-step explanation:

Assuming this complete problem: "The CEO of a software company is committed to expanding the proportion of highly qualified women in the organization's staff of salespersons. He believes that the proportion of women in similar sales positions across the country is less than 45%. Hoping to find support for his belief, he directs you to test

H0: p .45 vs H1: p < .45.

In doing so, you collect a random sample of 50 salespersons employed by his company, which is thought to be representative of sales staffs of competing organizations in the industry. The collected random sample of size 50 showed that only 18 were women.

Compute the p-value associated with this test. Place your answer, rounded to 4 decimal places, in the blank. For example, 0.3456 would be a legitimate entry."

1) Data given and notation

n=50 represent the random sample taken

X=18 represent the number of women in the sample selected

$\hat p=\frac{18}{50}=0.36$ estimated proportion of women in the sample

$p_o=0.45$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion of women is less than 0.45:  

Null hypothesis:$p\geq 0.45$  

Alternative hypothesis:$p < 0.45$  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

$z=\frac{0.36 -0.45}{\sqrt{\frac{0.45(1-0.45)}{50}}}=-1.279$  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

$p_v =P(z$

And we can use the following code to find it  "=NORM.DIST(-1.279,0,1,TRUE)"