We are given ,
µ = 26.6 , = 29.8 , s = 2.1 and n = 56
Part e) Test statistic :
t = =
t = 11.403
Part i) 95% confidence interval
Lower = - E
Upper = + E
E is margin of error = ; t is critical value follows t distribution with degrees of freedom (df) = n-1
We are given confidence level = 0.95
We can find critical value using excel function =TINV( α , d.f )
α = 1 - confidence level = 1 - 0.95 = 0.05
degrees of freedom (d.f) = n - 1 = 56-1 = 55
t =TINV(0.05,55)
t = 2.004
E is margin of error = =
E = 0.5624
Lower = - E = 29.8 - 0.5624
Lower = 29.24
Upper = + E = 29.8 + 0.5624
Upper = 30.36
#2) We are given x = 8 and n = 70 and P = 0.14
Part d) According to sampling distribution of proportion , sample proportion follows normal distribution with mean = P = 0.14 and standard deviation = = = 0.0415
P' - N ( 0.14 , 0.0415 )
Part e)
Test statistic : We have = x/n = 8/70 = 0.1143
Z = =
Z = -0.62
Part f)
This is left tail test , therefore p value would be the are to the left of -0.62
P value = P ( z ≤ -0.62 )
P- value = 0.2676
The mean age of De Anza College students in a previous term was 26.6 years old....
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