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 :-

$H_0: p=0.38\\\\H_a:p\neq0.38$

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 : $\hat{p}=\dfrac{522}{1558}\approx0.34$



The test statistic for proportion is given by :-

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

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