Given :An Internet survey was e-mailed to 6977 subjects randomly selected from an online group involved with ears. There were 1337 surveys returned.

To Find : Use a 0.01 significance level to test the claim that the return rate is less than 20%.

Solution:

n = 6977

x = 1337

We will use one sample proportion test

$\widehat{p}=\frac{x}{n}$

$\widehat{p}=\frac{1334}{6977}$

$\widehat{p}=0.1911$

We are given that the claim is the return rate is less than 20%.

$H_0:p=0.2\\H_a:p$

Formula of test statistic = $\frac{\widehat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

= $\frac{0.1911-0.2}{\sqrt{\frac{0.2(1-0.2)}{6977}}}$

= $−1.858$

Refer the z table

P(z<-1.85)=0.0332

Significance level = 0.01=α

Since p value > α

So, we accept the null hypothesis .

So,the claim that the return rate is less than 20% is false.

7 0

2 years ago

Find the value of s. Write your answer as a decimal number. 6 s 7 s =

Liono4ka [1.6K]

The value of s would be 6 or 7

8 0

2 years ago