Question

The quality control manager at a light bulb factory needs to estimate the mean life of a large shipment of light bulbs. The standard deviation is 98 hours. A random sample of 49 light bulbs indicated a sample mean life of 300 hours.

Suppose the standard deviation changes to 77 hours. What are your answers in​ (a) and​ (b)?

(Ans.) The 99​% confidence interval estimate would be from a lower limit of 271.7 hours to an upper limit of 328.3 hours.

Based on the sample data and a standard deviation of 77 ​hours, the manufacturer (does not have / has)-(which one) the right to state that the lightbulbs have a mean life of 350 hours. A mean of 350 hours is (less than 2 / more than 4)-(which one) standard errors (above / below)-(which one) the sample​ mean, so it is (highly unlikely / likely)-(which one) that the lightbulbs have a mean life of 350 hours.

Answer 1

z value at 99% = 2.576

CI = mean +/- z *(s/sqrt(n))
= 300 +/- 2.576 *(98/sqrt(49))
= (263.9 , 336.1 )

Lower limit = 263.9
Upper limit = 336.1

b)

mean = 300 , sd = 98
3 *sd = 3 * 98 = 294

350 > 294
the manufacturer does not have the right to state that the lightbulbs have a mean life of 350 hours.

c)

sd = 77

z value at 99% = 2.576

CI = mean +/- z *(s/sqrt(n))
= 300 +/- 2.576 *(77/sqrt(49))
= (271.7 , 328.3)

Lower limit = 271.7
Upper limit = 328.3

The 99% CI is 271.7 to 328.3 hours

Based on the sample​ data, the manufacturer does not have the right to state that the lightbulbs have a mean life of 350 hours. A mean of 350 hours is more than 3 standard errors above the sample​ mean, so it is highly unlikely that the lightbulbs have a mean life of 350 hours