Answer:
Step-by-step explanation:
Given that:
Mean = 3.34
sample size = 26
sample mean = 2.7
standard deviation = 1.17
level of significance = 0.10
The null hypothesis and the alternative hypothesis can be computed as follows:
[tex]\mathtt{H_o: \mu \geq 3.34} \\ \\ \mathtt{H_1: \mu < 3.34}[/tex]
degree of freedom = n - 1
degree of freedom = 26 -1
degree of freedom = 25
level of significance = 0.10
Since the alternative hypothesis contains <, then the test is left tailed
[tex]\mathtt{t_{\alpha, df} = t_{0.10, 25}}[/tex]
[tex]\mathtt{t_{0.10, 25}}[/tex] = - 1.316
The rejection region therefore consist of all values smaller than - 1.316, therefore ; reject [tex]H_o[/tex] if t < -1.316
The test statistics can be computed as follows:
[tex]t = \dfrac{X - \mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
[tex]t = \dfrac{2.7 - 3.34}{\dfrac{1.17}{\sqrt{26}}}[/tex]
[tex]t = \dfrac{-0.64}{\dfrac{1.17}{5.099}}[/tex]
t = - 2.789
Decision Rule: To reject the null hypothesis if the t test lies in the rejection region or less than the rejection region.
Conclusion: We reject the null hypothesis since t = (- 2.789) < -1.316. Then we conclude that the mean number of residents in the retirement community household is less than 3.34 persons.
