Question

Cell-phone passwords A consumer organization suspects that less than half of parents know their child’s cell-phone password. The Pew Research Center asked a random sample of parents if they knew their child’s cell-phone password. Of the $1060$parents surveyed, $551$reported that they knew the password. Explain why it isn’t necessary to carry out a significance test in this setting.

Short answer

Sample proportion is larger than$0.5$

Step-by-step solution

Given Information

It is given that $n=1060$

$x=551$

Claim is less than$50\%$

Calculation and Explanation

The claim can be either null or alternate hypothesis.

Null Hypothesis: $H_0:p=50\%=0.5$

Alternative Hypothesis: $H_1:p<0.5$

Sample proportion is calculated as:

$\hat{p}=\frac{x}{n}=\frac{551}{1060}=0.5198$

As claim is less than $0.5$and sample proportion is greater than $0.5$.

So, there is no proof provided by sample proportion that alternative hypothesis can be true.