Question

A scientist who studies teenage behavior was interested in determining if teenagers spend more time playing computer games then they did in the 1990s. In 1990s, the average amount of time spent playing computer games was 10.2 hours per week. Is the amount of time greater than that for this year? Ten students were surveyed and asked how many hours they spent playing video games. The test statistics is equal to 0.45.What is the p-value?greater than 0.10between 0.010 and 0.025less than 0.001between 0.001 and 0.005between 0.005 and 0.010between 0.025 and 0.05between 0.05 and 0.10

Answer 1

Answer:

greater than 0.10

Explanation:

The null hypothesis is:

$H_{0} = 10.2$

The alternate hypotesis is:

$H_{1} > 10.2$

Our test statistic is:

$t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the statistic, $\mu$ is the mean, $\sigma$ is the standard deviation and n is the size of the sample.

We have that:

$t = 0.45$

We are testing if X is greater than 0.45, so our pvalue is 1 subtracted by the pvalue of z = t = 0.45.

z = 0.45 has a pvalue of 0.6736

1 - 0.6735 = 0.3264

So our pvalue is 0.3264, which is greater than 0.10.

So the correct answer is:

greater than 0.10