In: Statistics and Probability
A random sample of 33 taxpayers claimed an average of $9,668 in medical expenses for the year. Assume the population standard deviation for these deductions was $2,415.
Construct confidence intervals to estimate the average deduction for the population with the levels of significance shown below.
a. |
1 % |
b. |
5 % |
c. |
20 % |
a. The confidence interval with a 1 % level of significance has a lower limit of $ and an upper limit of $
a)
confidence interval with a 1 % level of significance has a lower limit of 8585 and an upper limit of 10751
( please try 8583 to 10753 if above comes wrong due to rounding and revert)
b)
for 95 % CI value of z= | 1.960 | |||
margin of error E=z*std error = | 823.979 | |||
lower confidence bound=sample mean-margin of error= | 8844.02 | |||
Upper confidence bound=sample mean +margin of error= | 10491.98 |
confidence interval with a 5 % level of significance has a lower limit of 8844 and an upper limit of 10492
c)
for 80 % CI value of z= | 1.280 | |||
margin of error E=z*std error = | 538.109 | |||
lower confidence bound=sample mean-margin of error= | 9129.89 | |||
Upper confidence bound=sample mean +margin of error= | 10206.11 |
confidence interval with a 20% level of significance has a lower limit of 9130 and an upper limit of 10206
( please try 9129 to 10207 if above comes wrong due to rounding and revert)