21/7 unit rate and equation

In a recent Super Bowl, a TV network predicted that 50 % of the audience would express an interest in seeing one of its forthcom

coldgirl [10]

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:

$p_v =2*P(z$

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% .