Question

An online health and medicine company claims that about 50% of Internet users would go to an online sight first for information regarding health and medicine. A random sample of 1318 Internet users was asked where they will go for information the next time they need information about health or medicine; 606 said that they would use the Internet. Test the claim that less than 50% of Internet users get their health questions answered online.

Answer 1

The p-value of the test is 0.0018, which means that for a level of significance higher than this, there is sufficient evidence to claim that less than 50% of Internet users get their health questions answered online.

Explanation:

Test the claim that less than 50% of Internet users get their health questions answered online.

At the null hypothesis, we test that 50% of Internet users get their health questions answered online, that is:

`H_0: p = 0.5`

At the alternate hypothesis, we test that this proportion is less than 50%, that is:

`H_a: p < 0.5`

The test statistic is:

`z = (X - \mu)/((\sigma)/(√(n)))`

In which X is the sample mean,
`\mu` is the value tested at the null hypothesis,
`\sigma` is the standard deviation and n is the size of the sample.

0.5 is tested at the null hypothesis:

This means that
`\mu = 0.5, \sigma = √(0.5*0.5) = 0.5`

A random sample of 1318 Internet users was asked where they will go for information the next time they need information about health or medicine; 606 said that they would use the Internet.

This means that
`n = 1318, X = (606)/(1318) = 0.4598`

Test statistic:

`z = (X - \mu)/((\sigma)/(√(n)))`

`z = (0.4598 - 0.5)/((0.5)/(√(1318)))`

`z = -2.92`

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion below 0.4598, which is the p-value of z = -2.92.

Looking at the z-table, z = -2.92 has a p-value of 0.0018.

The p-value of the test is 0.0018, which means that for a level of significance higher than this, there is sufficient evidence to claim that less than 50% of Internet users get their health questions answered online.