Respuesta :
Answer:
The responses to these question can be defined as follows:
Step-by-step explanation:
[tex]n = 16\\\\ \bar{x}= 73.7\\\\\sigma = 12\\\\a = 0.05\ or\ a = 0.10\\\\H_{o} \ : \mu = 68\\\\H_{a} \ : \mu \neq 68\\\\a = 0.05\\\\[/tex]
critical values [tex]=\pm t0.025,15 = \pm 2.131\\\\[/tex]
[tex](n-1) = 15^{\circ}\\\\a = 0.10\\\\[/tex]
critical values[tex]= \pm t0.05,15 = \pm 1.753\\\\[/tex]
[tex](n-1) = 15^{\circ}\\\\[/tex]
Testing the statistic values:
[tex]t = \frac{x-\mu_{0}}{ \frac{s}{\sqrt{n}}}\\\\[/tex]
[tex]= \frac{73.7-68}{(\frac{12}{\sqrt{16}})}\\\\\ = \frac{5.7}{(\frac{12}{4})}\\\\ = \frac{5.7}{(3)}\\\\ = 1.9\\[/tex]
Test statistic ta [tex]= -1.90\ lies[/tex]
The critical values[tex]\pm t_{0.05,15} =\pm 1.753[/tex]
It is in the region of dismissal. We dismiss the 10% significant null hypothesis.
[tex]t_a = 1.90 \\\\df = 15\\\\a = 0.05\\\\p-value = 076831\\\\[/tex]
P - value is greater than the level of significance a= 0.05
Null hypothesis we don't reject. At a 95% level, the claim is justified.
[tex]t_a = 1.90\\\\ df = 15\\\\ a = 0.10\\\\p-value = 076831\\\\[/tex]
P - value below the meaning level a = 0.10, we reject the hypothesis null. At a level of 90% the claim is not justified.
