Question

A TV station claims that 38% of the 6:00 - 7:00 pm viewing audience watches its evening news program. A consumer group believes this is too high and plans to perform a test at the 5% significance level. Suppose a sample of 830 viewers from this time range contained 282 who regularly watch the TV station’s news program. Carry out the test and compute the p-value.

Answer 1

Answer:

$z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374$  

$p_v =P(Z<-2.374)=0.0088$  

So the p value obtained was a very low 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .  

Step-by-step explanation:

1) Data given and notation

n=830 represent the random sample taken

X=282 represent the people that regularly watch the TV station’s news program

$\hat p=\frac{282}{830}=0.340$ estimated proportion of people that regularly watch the TV station’s news program

$p_o=0.38$ 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 less than 0.38:  

Null hypothesis:$p\geq 0.38$  

Alternative hypothesis:$p < 0.38$  

When we conduct a proportion test we need to use the z statisitc, 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:  

$z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374$  

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 significance level provided $\alpha=0.05$. The next step would be calculate the p value for this test.  

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

$p_v =P(Z<-2.374)=0.0088$  

So the p value obtained was a very low 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 reject the null hypothesis, and we can said that at 5% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .