Answer
For example, an individual checking account at M=major US banks costs the banks more than 500 dollars per year. A recent random sample of 300 such checking accounts produced a mean annual cost of 600 dollars to major US banks. Assume the standard deviation of annual costs to major US banks of all such checking accounts is 30 dollars. Make a 99% confidence interval for the current mean annual cost to major US banks of all such checking accounts.
σ_ $\bar{x}\ $ = $ σ /\sqrt n $ = $ 30 /\sqrt 300 $ =1.7321
The 99% confidence interval for $\mu$ is
$\bar{x}\ $ ± zσ_${\bar{x}\ }$ = 600 ± 2.58 (1.7321) = 600 ± 4.4688 = 595.5312 to 604.4688
Suppose all the information given in that example remains the same. Now we decrease the confidence level to 95%. From the normal distribution table, z=1.96
$\bar{x}\ $ ± zσ_${\bar{x}\ }$ = 600 ± 1.96 (1.7321) = 600 ± 3.395 = 596.605 to 603.395
Comparing this confidence interval to the above example, we observe that the width of the 95% confidence interval is smaller than the 99% confidence interval.
Work Step by Step
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