Question

The number of events is 29​, the number of trials is 298​, the claimed population proportion is​ 0.10, and the significance level is 0.05. Use technology to identify the test statistic for this hypothesis​ test, rounding to two decimal places.

Answer 1

Answer:

$z=\frac{0.0973 -0.1}{\sqrt{\frac{0.1(1-0.1)}{298}}}=-0.155$  

$p_v =2*P(Z<-0.155)=0.877$  

And we can use excel to find the p value like this: "=2*NORM.DIST(-0.155;0;1;TRUE)"

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly different from 0.1 .  

Step-by-step explanation:

1) Data given and notation

n=298 represent the random sample taken

X=29 represent the events claimed

$\hat p=\frac{29}{298}=0.0973$ estimated proportion

$p_o=0.1$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

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 is 0.1 or no.:  

Null hypothesis:$p=0.1$  

Alternative hypothesis:$p \neq 0.1$  

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:  

$z=\frac{0.0973 -0.1}{\sqrt{\frac{0.1(1-0.1)}{298}}}=-0.155$  

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 bilateral test the p value would be:  

$p_v =2*P(Z<-0.155)=0.877$  

And we can use excel to find the p value like this: "=2*NORM.DIST(-0.155;0;1;TRUE)"

So the p value obtained was a very high value and using the significance level given $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is not significantly different from 0.1 .  

We can do the test also in R with the following code:

> prop.test(29,298,p=0.1,alternative = c("two.sided"),conf.level = 1-0.05,correct = FALSE)