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

z test statistic is -1.042 .

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
`\\eq` 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{(\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{(0.36(1-0.36))/(100) } }`

= -1.042

Therefore, the z test statistic is -1.042 .

Answer 2

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

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

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{(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{(0.41(1-0.41))/(100)}}=-1.017`