Answer to Question #308298 in Statistics and Probability for lanni

Question

a random sample of 429 college students was interviewed about a number of matters. use the results to construct confidence interval estimates of the population mean at the 99% level.

A. they reported that they had sent an average of $478.23 on textbooks during the previous semester , with a sample standard deviation of $15. 78

B. They also reported that they had visited the health clinic an average of 1.5 times a semester with a sample standard deviation of 0.3

C. on the average, the sample had missed 2.8 days of classes per semester because of illness with a sample standard deviation of 1.0

D. on the average, the sample had missed 3.5 days of classes per semester for reason other then illness, with a sample standard deviation of 1.5

please use excel

Answer 1

N=429, 99% confidence interval for the population mean is $μ±Z∙s/\sqrt{(N-1)}$

The critical value at 0.01 is 2.58

A. $\mu=478.23,s=15.78$

confidence interval = $478.23±2.58(15.78/\sqrt{(429-1)})$

$=478.23±1.97$

$=[476.26,480.2]$

Therefore the student spends between $476.26 and $480.20 on books.

B. $\mu=1.5,s=0.3$

confidence interval $=1.5±2.58(0.3/\sqrt{(429-1)})=1.5±0.04$

$=[1.46,1.54]$

Therefore the students visited the clinic between 1.46 and 1.54 times on average.

C. $\mu=2.8,s=1.0$

Confidence interval $=2.8±2.58(1.0/\sqrt{(429-1)})=2.8±0.12$

$=[2.67,2.93]$

Therefore the students missed classes per semester due to illness on the average between 2.67 and 2.93 times.

D. $\mu=3.5,s=1.5$

$=3.5±2.58(1.5/\sqrt{(429-1)})=3.5±0.19$

$=[3.31,3.69]$

Therefore the students missed classes per semester due to other reasons than illness on an average between 3.31 and 3.69.