In: Statistics and Probability
For a school, the principal expected that a student takes 45 minutes to travel to school on average. Suppose the time taken to travel to school by a student follows a normal distribution. 100 students are randomly selected from the school for interview and they took on average 44 minutes to travel to school with a standard deviation of 15 minutes. 6 of the selected students had lateness record in the last month.
Answer:
Sample Mean = 44
Std Dev = 15
n = 100
Std Error = 15/1001/2 = 1.5
a)
At alpha = 0.05,
Z Critical = 1.96
95% CI = Mean +/- ZCritical * Std Error = 44 +/- 1.96 * 1.5 = {41.06,46.94}
b)
Margin of Error = 0.8
Alpha = 0.06
ZCritical = 1.881
Std Dev = 10
Margin of Error = ZCritical * SE = ZCritical * Std Dev/n1/2
n = (1.881*10/0.8)2 = 552.84 ~ 553
So, Additional Sample = 553-100 = 453
c)
Alpha = 0.05,
N = 100
T Critical = +/- 1.984
Null and Alternate Hypothesis
H0: µ = 45
Ha: µ <> 45
Test Statistic
t = (44-45)/1.5 = - 0.67
Since the t value lies in the interval {-1.984,1.984} we fail to reject the null hypothesis ie true mean time taken to travel to school by student was 45minutes.
d)
Alpha = 0.075
Z Critical ~ 1.75
Proportion (p) = 6/100 = 0.06
Std Dev = {p*(1-p)/n}1/2 = 0.024
Hence 92.5% CI = 0.06 +/- 1.75*0.024 = {0.018,0.102}
NOTE : We are only allowed to answer 4 parts per question.