Question

One study revealed that a child under the age of 10 watches television 4.5 hours per day. A group of families from a certain comunity would like to beleive that their children watch television less than the national average. A random sample of 14 children from the community yielded a mean of 4.1 hours per day with standard deviation of 1.2. Test the appropriate hypothesis at the level of significance 0.01. Assume the viewing time is normally distributed and interpret your results.

Answer 1

`t=(4.1-4.5)/((1.2)/(√(14)))=-1.25`

The degrees of freedom are given by:

`df=n-1= 14-1=13`

And the p value would be:

`p_v =P(t_(13)<-1.25)=0.117`

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 4.5 minutes.

Explanation:

Information given

`\bar X=4.1` represent the sample mean

`s=1.2` represent the sample deviation

`n=14` sample size

`\mu_o =4.5` represent the value to test

`\alpha=0.01` represent the significance level

t would represent the statistic

`p_v` represent the p value

Hypothesis to test

We want to test if the true mean is less than 4.5, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 4.5`

Alternative hypothesis:
`\mu < 4.5`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Replacing the info given we got:

`t=(4.1-4.5)/((1.2)/(√(14)))=-1.25`

The degrees of freedom are given by:

`df=n-1= 14-1=13`

And the p value would be:

`p_v =P(t_(13)<-1.25)=0.117`

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 4.5 minutes.