The nominal resistance of a wire is 0.15 Ohm. Random testing of the wire stock yields the following resistance data: 0.148 0.147 0.151 0.146 0.151 0.148 0.147 0.152
0.151 0.148 0.149 0.147 0.146 0.149 0.151 0.147
Does the sign test yield the conclusion, at the 5% significance level, that the median resistance is less than 0.15 Ohm?

The median resistance of the wire is not small than 0,15 Ω at 0,05 of level of significance

Step-by-step explanation:

From data we get    x ( sample mean ) s  ( sample standard deviation) and n size of the sample

0.148 0.147 0.151 0.146 0.151 0.148 0.147 0.152

0.151 0.148 0.149 0.147 0.146 0.149 0.151 0.147

n = 16

x ≈ 0,1487

s = 0,00193

Test Hypothesis

1.-Null Hypothesis               H₀                x  =  μ₀  = 0,15

Alternative Hypothesis  Hₐ                 x < 0,15

We assume data follows normal distribution

as n = 16 we should use a t-student table

As the question is : "Is the median resistance less than 0,15 0hm " we must use  one-tail-test ( to the left)

Then:

2.-Significance level  α = 5 %      α = 0,05

degree of freedom   n = 16     df = n - 1    df = 15

From t-table we find t(c) = 1,7531      at the left is t(c) = - 1,7531

3.-t(s) = ( x - 0,15 ) / s / √n

t(s) = 0,1487 - 0,15 / 0,00193/√16

t(s) = - 0,0013 * 4 / 0,00193

t(s) = - 2,69

4.- Comparing t(s) and t(c)  

|t(s) | > |t(c)|       2,69 > 1,753

Then t(s) is in the rejection region

5.- We reject H₀  . At 95 % of confidence Interval

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

the answer is b. 60. each interior angle in 120, so the exterior angle must be 60 in order to add up to 180

step-by-step explanation: