Answer:
Step-by-step explanation:
The summary of the statistics given include:
population mean [tex]\mu[/tex] = 15
sample mean [tex]\oerline x[/tex] = 13.5
sample size n = 16
standard deviation s = 6
The level of significance ∝ = 0.10
The null and the alternative hypothesis can be computed as follows:
[tex]\mathtt{H_o: \mu = 15} \\ \\ \mathtt{H_1 : \mu \neq 15}[/tex]
Since this test is two tailed, the t- test can be calculated by using the formula:
[tex]t = \dfrac{\overline x - \mu}{\dfrac{\sigma }{\sqrt{n}}}[/tex]
[tex]t = \dfrac{13.5 - 15}{\dfrac{6}{\sqrt{16}}}[/tex]
[tex]t = \dfrac{- 1.5}{\dfrac{6}{4}}[/tex]
[tex]t = \dfrac{- 1.5\times 4}{6}}[/tex]
[tex]t = \dfrac{- 6.0}{6}}[/tex]
t = - 1
degree of freedom = n - 1
degree of freedom = 16 - 1
degree of freedom = 15
From the standard normal t probability distribution table, the p value when t = -1 at 0.10 level of significance, the p - value = 0.3332
Decision Rule: We fail to reject the null hypothesis since the p-value is greater than the level of significance at 0.10
Conclusion: Therefore, we can conclude that there is insufficient evidence at the 0.10 level of significance to conclude that the population mean μ is different than 15.
