Question

Find the critical value of t for a t-distribution with 30 degrees of freedom such that the area between -t and t is 99%

A student records the repair costs for 25 randomly selected computers from a local repair shop where he works. A sample mean of $216.53 and standard deviation of $15.86 are subsequently computed. Assume that the population distribution is approximately normal and s is unknown. Determine the 98% confidence interval for the mean repair cost for all computers repaired at the local shop by first calculating the margin of error, E.

Answer 1

solution

(A)

Degrees of freedom = df = 30

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df= t0.005,30 = +/-2.750 ( using student t table)

(B)Solution :

Given that,

Point estimate = sample mean =   =$216.53

Population standard deviation =    =$15.86

Sample size = n =25

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

Z/2 = Z0.01 = 2.326 ( Using z table    )

Margin of error = E = Z/2    * ( /n)

= 2.326 * ( 15.86 /   25)

= 7.3781
At 98% confidence interval estimate of the population mean
is,

- E < < + E

216.53 - 7.3781 <   < 216.53+ 7.3781

209.1519<   < 223.9081

( 209.1519, 223.9081 )