Question

A survey of 1,562 randomly selected adults showed that 522 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 35​% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial​ distribution, and assume a 0.05 significance level to complete parts​ (a) through​ (e). Is the test two-tailed, left-tailed, or right-tailed? Left-tailed test Two-tailed test Right tailed test What is the test statistic? (Round to two décimal places as needed.) What is the P-vahie?

Answer 1

a) We want to test the claim that 35​% of adults have heard of the new electronic reader, then the system of hypothesis are.:

Null hypothesis:
`p=0.35`

Alternative hypothesis:
`p \\eq 0.35`

And is a two tailed test

b)
`z=\frac{0.334 -0.35}{\sqrt{(0.35(1-0.35))/(1562)}}=-1.326`

c)
`p_v =2*P(z<-1.326)=0.184`

d) Null hypothesis:
`p=0.35`

e) Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.

Explanation:

Information provided

n=1562 represent the random sample selected

X=522 represent the people who have heard of a new electronic reader

`\hat p=(522)/(1562)=0.334` estimated proportion of people who have heard of a new electronic reader

`p_o=0.35` is the value to verify

`\alpha=0.05` represent the significance level

z would represent the statistic

`p_v` represent the p value

Part a

We want to test the claim that 35​% of adults have heard of the new electronic reader, then the system of hypothesis are.:

Null hypothesis:
`p=0.35`

Alternative hypothesis:
`p \\eq 0.35`

And is a two tailed test

Part b

The statistic for this case is given :

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

Replacing the info given we got:

`z=\frac{0.334 -0.35}{\sqrt{(0.35(1-0.35))/(1562)}}=-1.326`

Part c

We can calculate the p value using the laternative hypothesis with the following probability:

`p_v =2*P(z<-1.326)=0.184`

Part d

The null hypothesis for this case would be:

Null hypothesis:
`p=0.35`

Part e

The best conclusion for this case would be:

Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.