(1 point) If we randomly select groups of 20 batteries, measure how many hours t

Question

(1 point) If we randomly select groups of 20 batteries, measure how many hours they last, and then take the average of he 2o ileimes, we find that the average lifetimes are normaly distributled (a) We get a group of 20 batteries and find that their average lifetime is 106 hours with a standard deviation of 18 hours. What is the probability that we would get a group of 20 batteries with an average lifetime of 106 hours or more if the average lifetime of all batteries is 98 hours? Is this probability significant at the 0.1 level? (one tailed) A. Yes B. No C. Maybe (b) We get a group of 20 batteries and find that their average lifetime is 93 hours with a standard deviation of 18 hours. What is the probability that we would get a group of 20 batteries with an average lifetime of 93 hours or less if the average lifetime of all batteries is 98 hours? Is this probability significant at the 0.1 level? (one tailed) A. Yes B. No OC. Maybe (c) We get a group of 20 batteries and find that their average lifetime is 106 hours with a standard deviation of 18 hours. Based on this data, we can say with 95% confidence that the average lifetime of all batteries is between

Explanation / Answer

a)

P(X > 106)

mean = 98 , n = 20 , s = 18

z = ( x - mean)/ ( s/sqrt(n))

= ( 106 - 98) /( 18/sqrt(20))

= 1.9876

P(X > 106) = p(z > 1.9876) = 0.0234 by using standard normal table

yes, because probability is less than 0.1.

b)

P(X < 93)

mean = 98 , n = 20 , s = 18

z = ( x - mean)/ ( s/sqrt(n))

= ( 93 - 98) /( 18/sqrt(20))

= -1.2423

P(X < 93) = p(z < -1.2423) = 0.1071 = 0.11 by using standard normal table

N0 , because probability is greater than 0.1.

c)

By using z value

CI for 95%

n = 20

mean = 106

z-value of 95% CI = 1.9600

std. dev. = 18

SE = std.dev./sqrt(n) = 4.02492

ME = z*SE = 7.88870

Lower Limit = Mean - ME = 98.11130

Upper Limit = Mean + ME = 113.88870

95% CI (98.1113 , 113.8887 )

By using t value

CI for = 95%

n = 20

mean = 106

t-value of 95% CI = 2.0930

std. dev. = 18

SE = std.dev./sqrt(n) = 4.02492

ME = t*SE = 8.42426

Lower Limit = Mean - ME = 97.57574

Upper Limit = Mean + ME = 114.42426

95% CI (97.5757 , 114.4243 )

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