Question

A college chemistry instructor thinks the use of embedded tutors will improve the success rate in introductory chemistry courses. The passing rate for introductory chemistry is 68​%. During one​ semester, 200 students were enrolled in introductory chemistry courses with an embedded tutor. Of these 200 ​students, 152 passed the course. The instructor carried out a hypothesis test and found that the observed value of the test statistic was 2.43. The​ p-value associated with this test statistic is 0.0076. Explain the meaning of the​ p-value in this context. Based on this​ result, should the instructor believe the success rate has​ improved?

Answer 1

Answer:

$p_v = P(z>2.425) = 0.0076$

The p value is a value in order to reject or not the null hypothesis. If the p value is higher than the significance level we FAIL to reject the null hypothesis otherwise we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

X=152 represent the people passed the exam

$\hat p = \frac{152}{200}= 0.76$ estimated proportion of people report having texted while​ driving

$p_o = 0.68$ 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)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.68.:  

Null hypothesis:$p \leq 0.68$  

Alternative hypothesis:$p>0.68$  

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 $p_o$.

Calculate the statistic  

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

$z=\frac{0.76 -0.68}{\sqrt{\frac{0.68(1-0.68)}{200}}}=2.425$

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 right test the p value would be:  

$p_v = P(z>2.425) = 0.0076$

The p value is a value in order to reject or not the null hypothesis. If the p value is higher than the significance level we FAIL to reject the null hypothesis otherwise we have enough evidence to reject the null hypothesis.