Question

Based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station. What is the z test statistic?

Answer 1

Answer:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)  

$z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{100}}}=-1.017$  

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

$\hat p=0.36$ estimated proportion with the survey

$p_o=0.41$ 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 lower than 0.41.:  

Null hypothesis:$p\geq 0.41$  

Alternative hypothesis:$p < 0.41$  

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:  

$z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{100}}}=-1.017$  

Answer 2

Answer:

z test statistic is -1.042 .

Step-by-step explanation:

We are given that based on the Nielsen ratings, the local CBS affiliate claims its 11 p.m. newscast reaches 41% of the viewing audience in the area. In a survey of 100 viewers, 36% indicated that they watch the late evening news on this local CBS station.

Let Null Hypothesis, $H_0$ : p = 0.41 {means that % of the viewing audience in the area is 41%}

Alternate Hypothesis, $H_1$ : p $\neq$ 0.41 {means that % of the viewing audience in the area is different from 41%}

The z-test statistics we will use here is One sample proportion test ;

          T.S. = $\frac{\hat p - p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, p = % of the viewing audience based on the Nielsen ratings = 41%

       $\hat p$  = % of the viewing audience based on a survey of 100 viewers = 36%

       n = sample of viewers = 100

So, test statistics = $\frac{0.36 - 0.41}{\sqrt{\frac{0.36(1-0.36)}{100} } }$

                             = -1.042

Therefore, the z test statistic is -1.042 .