# DESPERATE for Stats homework help (10 points for best answer)!?

The supervisor of a production line believes that the average time to assemble an electronic component is 17 minutes. Assume that assembly time is normally distributed with a standard deviation of 3.8 minutes. The supervisor times the assembly of 20 components, and finds that the average time for completion was 17.85 minutes. Is there evidence that the average amount of time required to assemble a component is something other than 17 minutes?

We should reject at 9% level of significance if

test-statistic < -1.341

test-statistic > 1.392

| test-statistic | > 1.341

| test-statistic | > 1.695

If α= 0.09, what will be your conclusion?

not enough information to reach a decision

Do not reject H0

Reject H0

The p-value of the test is equal to

0.8414

0.1586

0.3171

Please show work/explain so I can understand this for my exam. THANK YOU!

Relevance
• Alan
Lv 7
2 weeks ago

Since you say different,

the 9 % alpha

mean from  alpha/2 to 1 - alpha/2

4.5 % to  95.5  leaving 91 % in the middle

so find P(z< Z) = 0.955

P(z<1.69) = .95449

P(z< 1.70) = 0.95543

so it is between 1.69 and 1.70

so reject it must least than -1.69 or greater than 1.69

| test-statistic | > 1.695

so for X = 17.85

Z =    (x-mean)/ (standard deviation/sqrt(N)

Z = (17.85-17) /  (3.8/sqrt(20))

Z = 0.85/ 0.8497058314 =  1.0003462005

since 1.0003 < 1.695 , you cannot reject the NULL Hypothesis

Do not reject H0

P(z<1.00)  read from the z-table =  0.84134

from this z-table

https://www.math.arizona.edu/~rsims/ma464/standard...