Respuesta :
Solution :
Given :
The sample mean = 21.3
Standard deviation = 3.2
The null hypothesis is : [tex]H_0: \mu =20[/tex]
The alternate hypothesis : [tex]H_a:\mu \neq20[/tex]
This is a two tailed test, for which a [tex]\text{t-test for one mean}[/tex] with an unknown population of a standard deviation is being used.
Now the significance level, [tex]\alpha = 0.1[/tex], as well as the critical value for a two tailed test is [tex]t_c = 1.833[/tex]
The rejection region is [tex]R = \{ t:|t| > 1.833 \}[/tex]
The t-statistic is computed as follows :
[tex]t=\frac{\overline x - \mu_0}{s/ \sqrt n}[/tex]
[tex]t=\frac{21.3-20}{3.2/ \sqrt{10}}[/tex]
= 1.285
Since it is observed that [tex]|t| = 1.285 \leq t_c=1.833[/tex], it is then concluded that [tex]\text{the null hypothesis is not rejected.}[/tex]
The p-value is p=0.231 and since p 0.231 ≥ 0.1, it is concluded that [tex]\text{the null hypothesis is not rejected.}[/tex]
Conclusion
Thus we concluded that [tex]\text{null hypothesis}[/tex] [tex]H_0[/tex] is not rejected. Therefore, the population mean [tex]\mu[/tex] is different than 20, at the 0.1 significance level.
