Answer to Question #147080 in Statistics and Probability for KNV

Question

(a) The life (in hours) of a 100 watt bulb is known to be normally distributed with standard deviation of 36 hours. A random sample of 15 bulbs yielded the following results:Life in hours 2,216 2,237 2,249 2,204 2,225 2,301 2,281 2,263 2,318 2,255 2,275 2,295 2,250 2,295 2,300.Construct a 95% two-sided confidence interval so that the actual mean of the life of bulbs falls within this interval.?

Answer 1

Mean (M) = (2,216 + 2,237 + 2,249 + 2,204 + 2,225 + 2,301 + 2,281 + 2,263 + 2,318 + 2,255 + 2,275 2,295 + 2,250 + 2,295 + 2,300)/15 = 2264,26

Based on the provided information (n = 15, α = 0.05) the critical z-value is Zc = 1.96

$\sigma$ = 36

Two-sided confidence interval:

$CI = (M - \frac{Z_c \times \sigma}{\sqrt{n}}, M + \frac{Z_c \times \sigma}{\sqrt{n}})$

$CI = (2264.26 - \frac{1.96 \times 36}{\sqrt{15}}, 2264.26 + \frac{1.96 \times 36}{\sqrt{15}})$

$CI = (2264.26 - 18.23, 2264.26 + 18.23)$

$CI = (2246.03, 2282.49)$