Exercise 2- Page 300 A simple random sample of 50 items from a population with σ = 6 resulted in a sample mean of 32. a. Provide a 90% confidence interval for the population mean: b. Provide a 95% confidence interval for the population mean
Trang 1I COMPULSORY HOMEWORK
Exercise 2- Page 300
A simple random sample of 50 items from a population with σ = 6 resulted in a sample mean of 32
a Provide a 90% confidence interval for the population mean:
b Provide a 95% confidence interval for the population mean
c Provide a 99% confidence interval for the population mean
SOLUTIONS
A simple random sample of 50 items from a population with σ = 6 resulted in a sample mean of 32
a Provide a 90% confidence interval for the population mean:
The general form of the 90% confidence interval for a population mean is
/2
6 32 1.645 (30.6;33.4)
50
x z
n
b Provide a 95% confidence interval for the population mean
The general form of the 95% confidence interval for a population mean is
/2
6 32 1.96 (30.34;33.66)
50
x z
n
c Provide a 99% confidence interval for the population mean
The general form of the 99% confidence interval for a population mean is
/2
6 32 2.575 (29.8;34.19)
50
x z
n
Notes: How to find the critical value (Zα/2)?
Step 1: define alpha
90% confidence interval means 1-α = 0.90, => α = 0.10 (α= significant level), so Zα/2 = ZA=Z0.05
Step 2: Ketch the normal curve
Trang 2Step 3: P(Z>Z0.05)=0.05
=> 1-P(Z< Z0.05) = 0.05
=> P(Z< Z0.05) = 0.95 Use table 1a, so Z5%= 1.645
Exercise 12- Page 308
Find the t-value(s) for each of the following cases
a Upper tail area of 0.025 with 12 degrees of freedom
b Lower tail area of 0.05 with 50 degrees of freedom
c Upper tail area of 0.01 with 30 degrees of freedom
d Where 90% of the area falls between these two t-values with 25 degrees of freedom
e Where 95% of the area falls between these two t-values with 45 degrees of freedom
SOLUTIONS
Find the t-value(s) for each of the following cases
a Upper tail area of 0.025 with 12 degrees of freedom
Using t-distribution table, with area in upper tail = 0.025 and df =n-1= 12
T 0.025 12 = 2.179.
Using Excel = TINV(2*0.025,12) = 2.1788
When using TINV in Excel, the probability implies the
significant level of 2-tails In case of calculating 1-tail t-score using TINV, you have to multiply the probability
(significance level) of one tail by 2.
Notes: * Always use the upper tail (P(t ≥ tα/2 n-1 ) for the T-table while always use the probability P(Z≤Z) in the left (the probability in the left, not the lower tail) for the Z-table.
Trang 3* In Microsoft Excel 2010, we have 2 functions for 1 tail and 2 tails as follows:
1 T.INV(probability, degree of freedom) for 1 tail
2 T.inv.2T(probability, degree of freedom) for 2 tails.
Remember T dot INV (not Tinv) and T dot Inv dot 2T (not Tinv2t)
Thus, in this case, you have to provide the exact probability (not multiplying by 2) you want in these 2 functions (T.INV and T.INV.2T).
= T.INV(0.975,12) = 2.17881
When using T.INV(p,df) and T.INV.2T(p,df), the probability p implies the total probability on the left (not the significant level of both 2 tails as in TINV)
b Lower tail area of 0.05 with 50 degrees of freedom
Using t-distribution table, with area in upper tail = 0.05 and df =n-1 = 50:
T 0.05 50 = 1.676
So the t-value at the lower tail is -1.676 because t-distribution is symmetric
Using Excel =TINV(2*0.01,50) = -1.6769
c Upper tail area of 0.01 with 30 degrees of freedom
Using t-distribution table, with area in upper tail = 0.01 and df = 30:
T 0.01 30 = 2.457
Using Excel = TINV(2*0.01,30) = 2.457
d Where 90% of the area falls between these two t-values with 25 degrees of
freedom
If there is 90% of the area between the 02 t-values, so, each tail area will be (1-0.9)/2
= 0.05
The area in the upper tail of 0.05, with df = 25, gives us t-value of 1.708
T 0.05 25 = 1.708
Excel: Tinv(2*0.05,25) = 1.7081
The other t-value is -1.708 (t-value from the lower tail)
e Where 95% of the area falls between these two t-values with 45 degrees of
freedom
If there is 95% of the area between the 02 t-values, so, each tail area will be (1-0.95)/2 =
0.025 The area in the upper tail of 0.025, with df = 45, gives us t-value of 2.014.
T 0.025 45 = 2.014
Excel: Tinv(2*0.05,45) = 2.0141
The other t-value is -2.014
Exercise 13- Page 308
The following sample data are from a normal population: 10, 8, 12, 15, 13, 11, 6, 5
Trang 4a What is the point estimate of the population mean?
b What is the point estimate of the population standard deviation?
c With 95% confidence, what is the margin of error for the estimation of the population mean?
d What is the 95% confidence interval for the population mean?
SOLUTIONS
The following sample data are from a normal population: 10, 8, 12, 15, 13, 11, 6, 5
a What is the point estimate of the population mean?
The point estimate of the population mean is the sample mean, which is 10
b What is the point estimate of the population standard deviation?
The point estimate of the population standard deviation is the sample standard
deviation s, which is 3.46
c With 95% confidence, what is the margin of error for the estimation of the
population mean?
/2 0.025
3.46
8
T 7
0.025 =2.365
Tinv(2*0.025,7)=2.365
d What is the 95% confidence interval for the population mean?
/2 0.025
3.46
8
x E x t x t
Exercise 27- Page 312
Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,000 and $45,000 Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired What is the planning value for the population standard deviation? How large a sample should be taken if the desired margin of error is
a $500?
b $200?
c $100?
d Would you recommend trying to obtain the $100 margin of error? Explain
SOLUTIONS
Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,000 and $45,000
Trang 5A Method 1: According to the Empirical rule (Textbook page 100), “approximately 95% of the data values will be within 2 standard deviations of the mean” (Exactly 1.96 standard deviations)
Because:
So,
Notes: Solving by this method, there is nothing to confuse between σ and σ xx
95% confidence interval estimate of the population mean, so Z α/2 =Z 0.025 = 1.96
This range is a 95% confidence interval estimate of the population mean annual starting salary, therefore, this range is 1.96 standard deviations from the population mean or (45,000 – 30,000)/2 = 7,500 from the mean
The planning value for the population STDEV is 7,500/1.96= 3,827= σ
B Method 2:
XX̅+ME=45,000
XX̅-ME=30,000
=> 2ME = 15,000
=> ME= 7,500
Denote ME= Margin of Error
However, Margin Error = Z α/2 δ xx = 7,500
Trang 6Thus, δ xx = ME/ Z α/2 = 7,500/1.96 = 3,827
a If the desired margin of error is $500, or
2 /2 /2
3,827
E
�
, so the sample size should be at least 225
b If the desired margin of error is $200, or
2 /2 /2
3,827
E
�
, so the sample size should be at least 1407
c If the desired margin of error is $100, or
2 /2 /2
3,827
E
�
, so the sample size should be at least 5625
In case we have a small population like 10,000 people, such a small margin of error of
$100 requiring 5625 items in the sample does not seem realistic It might raise the
associated costs and time Therefore, I would not recommend trying to obtain the $100 margin of error
In case we have a large number of items in the population such as 1,000,000 people, a sample size of 5625 is worth considering
Notes: E=Zα/2*σ/SQRT(n)
E when n
SQRT= Square root
Exercise 38- Page 317
According to Thomson Financial, through January 25, 2006, the majority of companies reporting profits had beaten estimates A sample of 162 companies showed 104 beat estimates, 29 matched estimates, and 29 fell short
a What is the point estimate of the proportion that fell short of estimates?
b Determine the margin of error and provide a 95% confidence interval for the
proportion that beat estimates
c How large a sample is needed if the desired margin of error is 05?
Trang 7According to Thomson Financial, through January 25, 2006, the majority of companies reporting profits had beaten estimates A sample of 162 companies showed 104 beat estimates, 29 matched estimates, and 29 fell short
a What is the point estimate of the proportion that fell short of estimates?
The point estimate of the proportion that fell short of estimates is the sample proportion
that fell short of estimates
29 0.1790 162
x p n
b Determine the margin of error and provide a 95% confidence interval for the
proportion that beat estimates
The point estimate of the proportion that beat estimates is the sample proportion that
beat estimates
104 0.6420 162
x p n
The margin of error and interval estimate with 95% confidence for the proportion that beat estimates is:
162
p
n
c How large a sample is needed if the desired margin of error is 05?
The planning value for p is taken from the sample of 162 companies: p* =
104/162=0.6420
/2
( ) (1 )
0.05
�
II ADDITIONAL EXCERCISES:
Exercise 4- Page 300
A 95% confidence interval for a population mean was reported to be 152 to 160 If σ =
15, what sample size was used in this study? Denote ME= Margin of Error
XX̅+ME=160
XX̅-ME=152
=> 2ME=8
=> ME=4
Trang 8The range for the 95% confidence interval is 160 – 152 = 8, which is two times the
margin of error Therefore, the margin of error is 8/2 = 4
We know that margin of error
/2
4
E
Exercise 5- Page 300
In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant, data were collected for a sample of 49 customers Assume a population
standard deviation of $5
a At 95% confidence, what is the margin of error?
/2
5
49
E z
n
b If the sample mean is $24.8, what is the 95% confidence interval for the
population mean?
/2
5 24.8 1.96 (23.4; 26.2)
49
x z
n
Exercise 6- Page 300 (CD file: Nielsen)
Nielsen Media Research conducted a study of household television viewing times during the 8pm to 11pm time period The data contained in the CD file named Nielsen are
consistent with the findings reported Based upon past studies the population standard deviation is assumed known with σ = 3.5 hours Develop a 95% confidence interval estimate of the mean television viewing time per week during the 8pm to 11pm time period
From the data collected, we know that the sample size n = 300
And we can compute the sample mean = 8.5
The 95% confidence interval estimate of the mean television viewing time per week during the 8pm to 11pm time period is
/2
3.5 8.5 1.96 (8.1;8.9)
300
x z
n
Exercise 16 – page 309
The mean number of hours of flying time for pilots at Continental Airlines is 49 hours per month Assume that this mean was based on actual flying times for a sample of 100 Continental pilots and that the sample standard deviation was 8.5 hours
a At 95% confidence, what is the margin of error?
Trang 9Given that level of confidence of 95%, the significant level is α = 1-0.95 = 0.05
The sample standard deviation is given as 8.5 hours
For a 95% confidence interval, we use Table 2 of Appendix B and n-1 =99 degree of freedom to obtain t100
0.025 = 1.984
Or use Excel = tinv(0.05,99)
The margin of error for 95% confidence interval is:
/2 0.025
8.5
100
b What is the 95% confidence interval estimate of the population mean flying time for the pilots?
The 95% confidence interval is 49 ± margin of error = 49 ± 1.6866 = (47.31; 50.69)
c The mean number of hours of flying time for pilots at United Airline is 36 hours per month Use your results from part (b) to discuss differences between the flying times for the pilots at the two airlines The Wall Street Journal reported United Airlines as having the highest labor cost among all airlines Does the information in this exercise provide insight as to why United Airlines might expect higher labor costs?
The information in this exercise shows that Continental Airlines’ pilots work more
efficiently than United Airlines’ pilots as they fly an average of 49 hours compared with that of 36 hours for United Airlines’ pilots This fact implied that United Airlines might expect higher labor costs
Conclusion: Continental: More efficient
Exercise 21- Page 310 (CD file: Alcohol)
Consumption of alcoholic beverages by young women of drinking age has been
increasing in the UK, the US, and Europe Data (annual consumption in liters) consistent with the findings reported in The Wall Street Journal article are shown for a sample of 20 European young women (file 8.8.Alcohol)
Assuming the population is roughly symmetric, construct a 95% confidence interval for the mean annual consumption of alcoholic beverages by European young women
As the population is assumed to be roughly symmetric, the sampling distribution of the sample mean should be approximated by t-distribution with degree of freedom of 19 because the population standard deviation is unknown
Given that level of confidence of 95%, the significant level is α = 1-0.95 = 0.05
Trang 10The sample mean is calculated from the sample data: x = 130
The sample standard deviation is calculated from the sample data and: s = 65.39 liters For a 95% confidence interval, we use Table 2 of Appendix B and n-1 =19 degree of freedom to obtain t0.025 = 2.093
Or use Excel = tinv(0.05,19)
The margin of error for 95% confidence interval is:
/2 0.025
65.39
20
The 95% confidence interval for the mean annual consumption of alcoholic beverages by European young women is:
130 ± margin of error = 130 ± 30.60 = (99.39; 160.60)
Exercise 28- Page 312
An online survey by ShareBuilder, a retirement plan provider, and Harris Interactive reported that 60% of female business owners are not confident they are saving enough for retirement Suppose we would like to do a follow-up study to determine how much female business owners are saving each year toward retirement and want to use $100 as the desired margin of error for an interval estimate of the population mean Use $1100 as
a planning value for the standard deviation and recommend a sample size for each of the following situations
a A 90% confidence interval is desired for the mean amount saved.
If we want a margin of error of $100 and a 90% confidence level, the sample size needs
to be at least
100
z n E
�
b A 95% confidence interval is desired for the mean amount saved.
If we want a margin of error of $100 and a 95% confidence level, the sample size
needs to be at least
100
z n E
�
c A 99% confidence interval is desired for the mean amount saved.
If we want a margin of error of $100 and a 99% confidence level, the sample size
needs to be at least
100
z n E
�
d When the desired margin of error is set, what happens to the sample size as the
confidence level is increased? Would you recommend using a 99% confidence interval in this case? Discuss
Trang 11When the desired margin of error is set (fixed), the sample size needs to be larger if the confidence level is increased Requiring a 99% confidence interval in this case would require a sample size of at least 803 It might be very costly and inconvenient trying to achieve that sample size For this type of research, it is not very necessary to achieve a 99% confidence level
Exercise 31- Page 316
A simple random sample of 400 individuals provides 100 Yes responses
a What is the point estimate of the proportion of the population that would provide Yes responses?
The point estimate of the proportion of the population that would provide Yes responses
is the sample proportion
100 0.25 400
x p n
b What is your estimate of the standard error of the proportion?
The standard error of the proportion:
0.022 400
p
n
c Compute the 95% confidence interval for the population proportion.
In this situation, we have n.p = 100 >5 and n.(1-p) = 300>5, then the sample size is large enough to assume that the sampling distribution of the sample proportion is normal With 95% confidence, or α = 0.05, the interval estimate for the population proportion
400
p
n
Exercise 36- Page 317
According to statistics reported on CNBC, a surprising number of motor vehicles are not covered by insurance Sample results, consistent with the CNBC report, showed 46 of
200 vehicles were not covered by insurance
a What is the point estimate of the proportion of vehicles not covered by insurance?
The point estimate of the proportion of vehicles not covered by insurance is the sample
proportion
46 0.23 200
x p
n
b Develop a 95% confidence interval for the population proportion.
In this situation, we have n.p = 46 >5 and n.(1-p) = 154>5, then the sample size is large enough to assume that the sampling distribution of the sample proportion is normal