P(at least one wrong) = 1- P(all correct) =1-.25^6=1-1/4096=4095/4096. This assumes that the answers are picked at random. This kind of question is always the complement of an extreme binomial outcome.

4 0

2 years ago

In order to justify this claim for the Ferengi a sample of 16 coils is taken. For this sample set, the mean warp phase flux is f

Kay [80]

Answer:

The responses to these question can be defined as follows:

Step-by-step explanation:

$n = 16\\\\ \bar{x}= 73.7\\\\\sigma = 12\\\\a = 0.05\ or\ a = 0.10\\\\H_{o} \ : \mu = 68\\\\H_{a} \ : \mu \neq 68\\\\a = 0.05\\\\$

critical values $=\pm t0.025,15 = \pm 2.131\\\\$

$(n-1) = 15^{\circ}\\\\a = 0.10\\\\$

critical values $= \pm t0.05,15 = \pm 1.753\\\\$

$(n-1) = 15^{\circ}\\\\$

Testing the statistic values:

$t = \frac{x-\mu_{0}}{ \frac{s}{\sqrt{n}}}\\\\$

$= \frac{73.7-68}{(\frac{12}{\sqrt{16}})}\\\\\ = \frac{5.7}{(\frac{12}{4})}\\\\ = \frac{5.7}{(3)}\\\\ = 1.9\\$

Test statistic ta $= -1.90\ lies$

The critical values $\pm t_{0.05,15} =\pm 1.753$

It is in the region of dismissal. We dismiss the 10% significant null hypothesis.

$t_a = 1.90 \\\\df = 15\\\\a = 0.05\\\\p-value = 076831\\\\$

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.

$t_a = 1.90\\\\ df = 15\\\\ a = 0.10\\\\p-value = 076831\\\\$

P - value below the meaning level a = 0.10, we reject the hypothesis null. At a level of 90% the claim is not justified.

3 0

2 years ago