Download Answers Ch 9-10 Mixed File

STAT202 QUIZ 1
22.3.2011
1) A recent study by the Los Angeles Taxi Drivers Association showed that the mean fare
charged for service from Hermosa Beach to Los Angeles International Airport is $18
and the standard deviation is $3.50. We select a sample of 15 fares
a) What is the probability that the sample mean is between $17.00 and $20.00?
Solution:
z1 
X 
17  18
1


 1.11 ; p = 0.3665
 / n 3.5 / 15 0.9037
and z2 
20  18
 2.21 ; p = 0.4864.
0.9037
Therefore, p(17< x < 20) = 0.3665 + 0.4864 = 0.8529
b) What must you assume to make the above calculation?
Since population standard deviation is known, we assume that population is
normally distributed so that we can use z-distribution for small sample, n ≤ 30.
2) A recent survey of 50 executives who lost their job showed that it took 26 weeks for
them to find another job. The standard deviation of the sample was 6.2 weeks.
Construct a 95% confidence interval for the population mean. Is it reasonable to
conclude that the population mean is 28 weeks?
Solution:
s
n
 6.2 
26  1.96
 → 26  1.72
 50 
So, 95% confidence interval is 24.3 and 27.7
It is not reasonable to conclude that population mean is 28 because, it falls outside the
95% confidence interval.
X  z.
3) The Policy of the Suburban Transit Authority is to add a bus root if more than 55% of
the potential citizens indicate they would use the particular root. A sample of 70
citizens revealed that 42 would use the proposed root. Will the Suburban Transit
Authority add this bus root? Use the 0.05 level of significance.
Solution:
Step I:
H o :   0.55
H1 :  > 0.55
p
42
 0 .6
70
Step II: level of significance is 0.05. Critical z = 1.65.
Step III: we use z-distribution since both n(π) > 5 and n (1-π) > 5 . Reject null if
computed z > 1.65
Step IV:
z
p 
0.60  0.55
0.05


 0.841
 (1   )
0.55(0.45) 0.0595
n
70
Step V:
Since 0.841 < 1.65 we cannot reject the null hypothesis and conclude that Suburban
Transit Authority will not add the bus root.