In a sample of 36 suitcases , the avg is 45 lbs. Standard deviation is 8.2....

In a sample of 36 suitcases , the avg is 45 lbs. Standard deviation is 8.2. Construct at 90% confidence interval for the mean of the weight of all suticases

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Expert Solution

Solution :

Given that,

= 45

= 8.2

n = 36

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z_/2* ( / $\sqrt$ n)

=1.645 * (8.2 / $\sqrt$ 36)

= 2.2482

At 90% confidence interval estimate of the population mean is,

- E < < + E

45 - 2.2482 < < 45 + 2.2482

42.7518 < < 47.2482

(42.7518 , 47.2482)

orchestra answered 1 year ago

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