Question

A random sample of 57 children with working mothers showed that they were absent from school an average of 5.3 days per term with a standard deviation of 1.8 days. If days absent follows a normal distribution, determine the critical value for the test statistic (z or t) needed to build a 98% confidence interval for the average number of days absent per term for all children. State the positive value only exactly as found in the table (3 decimal places).

Answer 1

Given : confidence level=98 %

Mean=5.3 days

Standard deviation=1.8 days

alpha=1-confidence level/100

=1-(98/100)

=0.02

critical probability=1-alpha/2=1-0.02/2=0.98/2=0.49

Look for this value(0.49) in the Z table ,we get

Zcritical =2.3+0.03 =2.33

( Left end of the value and top most value from 0.49 in the z table gives the critical value)