Question

According to the Census Bureau, 3.39 people reside in the typical American household. A sample of 26 households in Arizona retirement communities showed the mean number of residents per household was 2.73 residents. The standard deviation of this sample was 1.22 residents. At the .10 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.39 persons?

Answer 1

`t=(2.73-3.39)/((1.22)/(√(26)))=-2.758`

`df=n-1=26-1=25`

`p_v =P(t_((25))<-2.758)=0.0054`

Since the p value is lower than the significance level 0.1 we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is significanlty lower than 3.39 personas at 10% of significance.

Explanation:

Data given and notation

`\bar X=2.73` represent the sample mean

`s=1.22` represent the sample standard deviation

`n=26` sample size

`\mu_o =3.39` represent the value that we want to test

`\alpha=0.1` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is less than 3.39 persons, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 3.39`

Alternative hypothesis:
`\mu < 3.39`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(2.73-3.39)/((1.22)/(√(26)))=-2.758`

P-value

The degreed of freedom are given by:

`df=n-1=26-1=25`

Since is a one sided lower test the p value would be:

`p_v =P(t_((25))<-2.758)=0.0054`

Conclusion

Since the p value is lower than the significance level 0.1 we have enough evidence to reject the null hypothesis, and we can conclude that the true mean is significanlty lower than 3.39 personas at 10% of significance.