1. A simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is 15.a. Compute the 95 % confidence interval for the population mean. b. Assume that the sample mean was obtained from a sample of 120 items. Provide a 99% confidence interval for the population mean.

a) ( 76.20, 83.80)

b) ( 76.47, 83.53)

Step-by-step explanation:

a)

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

Given that;

Mean x = 80

Standard deviation r = 15

Number of samples n = 60

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values we have;

80 +/-1.96(15/√60)

80+/-1.96(1.936491673103)

80+/- 3.80

= ( 76.20, 83.80)

Therefore at 95% confidence interval = ( 76.20, 83.80)

b)

Mean x = 80

Standard deviation r = 15

Number of samples n = 120

Confidence interval = 99%

z(at 99% confidence) = 2.58

Substituting the values we have;

80 +/-2.58(15/√120)

80+/-2.58(1.369306393762)

80+/- 3.53

= ( 76.47, 83.53)

Therefore at 99% confidence interval = ( 76.47, 83.53)

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step-by-step explanation:

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