Question

In a recent Super Bowl, a TV network predicted that 50 % of the audience would express an interest in seeing one of its forthcoming television shows. The network ran commercials for these shows during the Super Bowl. The day after the Super Bowl, and Advertising Group sampled 106 people who saw the commercials and found that 48 of them said they would watch one of the television shows.Suppose you are have the following null and alternative hypotheses for a test you are running:H0:p=0.5Ha:p≠0.5Calculate the test statistic, rounded to 3 decimal placesz=

Answer 1

Answer:

z= -0.968

We can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that  they would watch one of the television shows not differs from 0.5 or 50% .  

Step-by-step explanation:

1) Data given and notation n  

n=106 represent the random sample taken

X=48 represent the people who says that  they would watch one of the television shows.

$\hat p=\frac{48}{106}=0.453$ estimated proportion of people who says that  they would watch one of the television shows.

$p_o=0.5$ 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 50% of people who says that  they would watch one of the television shows.:  

Null hypothesis:$p=0.5$  

Alternative hypothesis:$p \neq 0.5$  

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:  

$z=\frac{0.453 -0.5}{\sqrt{\frac{0.5(1-0.5)}{106}}}=-0.968$  

4) Statistical decision  

P value method or p value approach . "This method consists on 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 is not provided, but we can assume $\alpha=0.05$. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

$p_v =2*P(z<-0.968)=0.333$  

So based on the p value obtained and using the significance level assumed $\alpha=0.05$ we have $p_v>\alpha$ so we can conclude that we fail to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who says that  they would watch one of the television shows not differs from 0.5 or 50% .