Question

A survey of 1 comma 558 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 38​% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial​ distribution, and assume a 0.01 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

Answer with explanation:

Let p be the proportion of adults have heard of the new electronic reader.

Given claim : The accompanying technology display results from a test of the claim that 38​% of adults have heard of the new electronic reader.

i.e.
`p=0.38`

Then , the set of hypothesis will be :-

`H_0: p=0.38\\\\H_a:p\\eq0.38`

Since, the alternative hypothesis is two tailed , so the test is two-tailed test.

Also, it is given that the sample size :
`n=1558`

Number of adults showed that they have heard of a new electronic reader=522

So the sample proportion for adults have heard of the new electronic reader :
`\hat{p}=(522)/(1558)\approx0.34`



The test statistic for proportion is given by :-

`z=\frac{\hat{p}-p}{\sqrt{(p(1-p))/(n)}}\\\\\\\Rightarrow\ z=\frac{0.34-0.38}{\sqrt{(0.38(1-0.38))/(1558)}}\\\\\\\Rightarrow\ z=-3.25279036541\approx-3.25`

By using standard normal distribution table , the P-value for two tailed test corresponds to the obtained z-value =
`=0.0011541`