The quality-control manager at a compact flourescent light bulb factory wants to test the claim that the mean life of a large shipment of cfls is equal to 6500 hours. the population standard deviation is 500 hours. a random sample of 50 cfls indicates a sample mean life of 6,700 hours and sample standard deviation is 600 hours. Critical values, where P(Z > Z) - a and P(t >) - a 2016-1282 20.06-1645 20.02 -1.960 10.10. - 1.299 10.06.45 - 1677 foto - 2010 1.115 pt.
a. At the 0.05 level of significance, is there evidence that the mean life is different from 6,500 hours?
b. Compute the p value and interpret its meaning.
c. Construct a 95% confidence interval estimate of the population mean life of the CFLs.
d. Compare the results of (a) and (c). What conclusions do you reach?

a. At the 0.05 level of significance,  there is evidence that the mean life is different from 6,500 hours.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. CI [6583.336 ,6816.336]

d.  The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

Step-by-step explanation:

Population mean = u= 6500 hours.

Population standard deviation = σ=500 hours.

Sample size =n= 50

Sample mean =x`=  6,700 hours

Sample standard deviation=s=  600 hours.

Critical values, where P(Z > Z) =∝ and P(t >) =∝

Z(0.10)=1.282  

Z(0.05)=1.645  

Z(0.025)=1.960  

t(0.01)(49)= 1.299

t(0.05)= 1.677  

t(0.025,49)=2.010

Let the null and alternate hypotheses be

H0: u = 6500 against the claim Ha: u ≠ 6500

Applying Z test

Z= x`- u/ s/√n

z= 6700-6500/500/√50

Z= 200/70.7113

z= 2.82=2.82

Applying  t test

t= x`- u /s/√n

t= 6700-6500/600/√50

t= 2.82

a. At the 0.05 level of significance,  there is evidence that the mean life is different from 6,500 hours.

Yes we reject H0  for z- test as it falls in the critical region,at the 0.05 level of significance, z=2.82 > z∝=1.645

For t test  we reject H0   as it falls in the critical region,at the 0.05 level of significance, t=2.82 > t∝=1.677 with n-1 = 50-1 = 49 degrees of freedom.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. The 95 % confidence interval of the population mean life is estimated by

x` ±  z∝/2  (σ/√n )

6700± 1.645 (500/√50)

6700±116.336

6583.336 ,6816.336

d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

the most boring ever written

step-by-step explanation:

i am writing something. yes, i plan to make it the most boring thing ever written. i go to the store. a car is parked. many cars are parked or moving. some are blue. some are tan. they have windows. in the store, there are items for sale. these include such things as soap, detergent, magazines, and lettuce. you can enhance your life with these products. soap can be used for bathing, be it in a bathtub or in a shower. apply the soap to your body and rinse. detergent is used to wash clothes. place your dirty clothes into a washing machine and add some detergent as directed on the box. select the appropriate settings on your washing machine and you should be ready to begin. magazines are stapled reading material made with glossy paper, and they cover a wide variety of topics, ranging from news and politics to business and stock market information. some magazines are concerned with more recreational topics, like sports card collecting or different kinds of hairstyles. lettuce is a vegetable.

the answer is y = 11.54

hope this : )