Question

Claim Most adults would erase all of their personal information online if they could. A software firm survey of 693 randomly selected adults showed that 58% of them would ease all of their personal information onif they could. Find the value of the test statistic

Answer 1

Step-by-step explanation

Let's see the facts:

Number of adults = 693

-58% would erase the personal information.

-100%-58% = 42% would not erase their personal information.

n= 693

p=58% = 0.58

As we can see, 58% represents the proportion of a sample and thus we are making a claim about a population proportion p in the hypotheses.

It says "most of the adults", so it would be p>0.5

The null hypotheses states that the population proportion is equal to the value mentioned in the claim.

H0 : p=0.5

Then, q=1-p = 1-0.5 = 0.5

Now, the test statistic will be:

`\frac{\vec{p}-p}{\sqrt[]{(pq)/(n)}}=\frac{0.58-0.5}{\sqrt[]{(0.5\cdot0.5)/(693)}}=\frac{0.08}{\sqrt[]{(0.25)/(693)}}=\frac{0.08}{\sqrt[]{(1)/(2772)}}=4.211`

Answer: the test statistic is equal to 4.211