Exercise 1
Which of the following statements about the mean is not always correct?
a) The sum of the deviations from the mean is zero
b) Half of the observations are on either side of the mean
c) The mean is a measure of the middle (centre) of a distribution
d) The value of the mean times the number of the observations equals the sum of all of the observations
Answer:
Exercise 2
In a probability distribution, the proportion of the total area which must be to the left of the median is:
a) Exactly 0.50
b) Less than 0.50 if the distribution is skewed to the left
c) More than 0.50 if the distribution is skewed to the left
d) Between 0.25 and 0.60 if the distribution is symmetric and unimodal
Answer:
Exercise 3
In a large class in statistics, the final examination grades have a mean of 67.4 and a standard deviation of 12. Assuming that the distribution of these grades is normal, find the following quantities. (You need to use a normal table or Excel in this exercise. See instructions on next page.)
a) the percentage of grades that should exceed 85;
b) the percentage of grades that is less than 45;
c) the number of passes (pass mark is 50) in a class of 180;
d) the lowest distinction mark if the highest 8% of grades are to be regarded as distinctions.
Exercise 4
An experiment involves selecting a random sample of 256 middle managers for study. One item of interest is annual income. The sample mean is computed to be £35,420, and the sample standard deviation is £2,050.
a) What is the estimated mean income of all middle managers (the population)?
b) Give a 95 percent confidence interval (rounded to the nearest £10) for your estimate of the mean income. Do you have to make any assumptions?
c) Interpret the meaning of the confidence interval.
Exercise 5
Please discuss or write down the answers to the following questions.
a) If you collect 4 times more data, how much narrower will your confidence interval (CI) be? Same question for collecting 100 times more data.
b) Assume you work for a manager who says one day “I got the budget to collect twice as much data; that’s great because our estimates will be twice as precise.” Is anything wrong with his statement?
c) Your manager says “Let’s just calculate our CIs with 90% coverage probability instead of 95%; this will make the CIs narrower.” Is she right or wrong? Your manager adds: “We get better precision this way.” What is the manager’s misconception?
Exercise 6
The processing times for a particular product follow a normal distribution with an assumed mean of 40 minutes, and a standard deviation of 5 minutes. Suppose we want to test that the process is under control using a 5% significance test of an observed sample mean.
A sample of 100 units yields an average processing time of 41.2 minutes.
a) Is the machine producing at its expected speed? (test at 5% significance level)
b) Is the assumption that the processing times follow a normal distribution necessary?
X
N(, )
The Normal Table
The normal table (see next page) gives the area under the curve of a normal distribution. Instead of listing all possible normal distributions (an impossible task!), the table is designed for the standard normal distribution that has a mean 0 and a standard deviation 1, N(0,1).
If we have a normal distribution N(,) and want to find the area under the curve above a number X (the shaded area above), we need to go through the two following steps:
1. Find how many standard deviations X is from the mean , i.e. the Z-value for X:
X Z
2. Go to the normal table and look up the area that corresponds to Z.
For example, if we have a distribution N(2, 0.5) and want to find the area above 2.23, we find the Z-value:
46 .05 .02 23 .2 Z
Hence, 2.23 is 0.46 standard deviations away from the mean, 2. Next we look up the area that corresponds to the Z-value 0.46 in the normal table. To locate 0.46 in the table we first locate 0.4 in the first column, then we locate 0.06 in the first row. The area is the value in the intersecting cell of the row and column, 0.3228, i.e., 32.28%.
Note that the table is only for positive Z-values. In order to determine the area under the curve for a negative Z-value we still can use the table by utilizing the fact that the normal distribution is symmetric.
Normal Distribution in Excel
We can find the area under the normal curve in Excel with the function:
NORM.DIST(X,mean,stdev,true)
where “X” determines the area of interest, “mean” is the mean of the distribution we are working with, “stdev” is the standard deviation and “true” is to indicate that we want the area under the curve (not the height).
Note that this function gives the area below X (not above as in the normal table). Hence in order to get the area above X (as listed in the normal table) we use:
1-NORM.DIST(X,mean,stdev,true)
N(, )
X
| The Normal Table Z = (X – )/ | 0.00 | 0.01 | 0.02 | 0.03 | 0.04 | 0.05 | 0.06 | 0.07 | 0.08 | 0.09 |
| 0.00 | 0.5000 | 0.4960 | 0.4920 | 0.4880 | 0.4840 | 0.4801 | 0.4761 | 0.4721 | 0.4681 | 0.4641 |
| 0.10 | 0.4602 | 0.4562 | 0.4522 | 0.4483 | 0.4443 | 0.4404 | 0.4364 | 0.4325 | 0.4286 | 0.4247 |
| 0.20 | 0.4207 | 0.4168 | 0.4129 | 0.4090 | 0.4052 | 0.4013 | 0.3974 | 0.3936 | 0.3897 | 0.3859 |
| 0.30 | 0.3821 | 0.3783 | 0.3745 | 0.3707 | 0.3669 | 0.3632 | 0.3594 | 0.3557 | 0.3520 | 0.3483 |
| 0.40 | 0.3446 | 0.3409 | 0.3372 | 0.3336 | 0.3300 | 0.3264 | 0.3228 | 0.3192 | 0.3156 | 0.3121 |
| 0.50 | 0.3085 | 0.3050 | 0.3015 | 0.2981 | 0.2946 | 0.2912 | 0.2877 | 0.2843 | 0.2810 | 0.2776 |
| 0.60 | 0.2743 | 0.2709 | 0.2676 | 0.2643 | 0.2611 | 0.2578 | 0.2546 | 0.2514 | 0.2483 | 0.2451 |
| 0.70 | 0.2420 | 0.2389 | 0.2358 | 0.2327 | 0.2296 | 0.2266 | 0.2236 | 0.2206 | 0.2177 | 0.2148 |
| 0.80 | 0.2119 | 0.2090 | 0.2061 | 0.2033 | 0.2005 | 0.1977 | 0.1949 | 0.1922 | 0.1894 | 0.1867 |
| 0.90 | 0.1841 | 0.1814 | 0.1788 | 0.1762 | 0.1736 | 0.1711 | 0.1685 | 0.1660 | 0.1635 | 0.1611 |
| 1.00 | 0.1587 | 0.1562 | 0.1539 | 0.1515 | 0.1492 | 0.1469 | 0.1446 | 0.1423 | 0.1401 | 0.1379 |
| 1.10 | 0.1357 | 0.1335 | 0.1314 | 0.1292 | 0.1271 | 0.1251 | 0.1230 | 0.1210 | 0.1190 | 0.1170 |
| 1.20 | 0.1151 | 0.1131 | 0.1112 | 0.1093 | 0.1075 | 0.1056 | 0.1038 | 0.1020 | 0.1003 | 0.0985 |
| 1.30 | 0.0968 | 0.0951 | 0.0934 | 0.0918 | 0.0901 | 0.0885 | 0.0869 | 0.0853 | 0.0838 | 0.0823 |
| 1.40 | 0.0808 | 0.0793 | 0.0778 | 0.0764 | 0.0749 | 0.0735 | 0.0721 | 0.0708 | 0.0694 | 0.0681 |
| 1.50 | 0.0668 | 0.0655 | 0.0643 | 0.0630 | 0.0618 | 0.0606 | 0.0594 | 0.0582 | 0.0571 | 0.0559 |
| 1.60 | 0.0548 | 0.0537 | 0.0526 | 0.0516 | 0.0505 | 0.0495 | 0.0485 | 0.0475 | 0.0465 | 0.0455 |
| 1.70 | 0.0446 | 0.0436 | 0.0427 | 0.0418 | 0.0409 | 0.0401 | 0.0392 | 0.0384 | 0.0375 | 0.0367 |
| 1.80 | 0.0359 | 0.0351 | 0.0344 | 0.0336 | 0.0329 | 0.0322 | 0.0314 | 0.0307 | 0.0301 | 0.0294 |
| 1.90 | 0.0287 | 0.0281 | 0.0274 | 0.0268 | 0.0262 | 0.0256 | 0.0250 | 0.0244 | 0.0239 | 0.0233 |
| 2.00 | 0.0228 | 0.0222 | 0.0217 | 0.0212 | 0.0207 | 0.0202 | 0.0197 | 0.0192 | 0.0188 | 0.0183 |
| 2.10 | 0.0179 | 0.0174 | 0.0170 | 0.0166 | 0.0162 | 0.0158 | 0.0154 | 0.0150 | 0.0146 | 0.0143 |
| 2.20 | 0.0139 | 0.0136 | 0.0132 | 0.0129 | 0.0125 | 0.0122 | 0.0119 | 0.0116 | 0.0113 | 0.0110 |
| 2.30 | 0.0107 | 0.0104 | 0.0102 | 0.0099 | 0.0096 | 0.0094 | 0.0091 | 0.0089 | 0.0087 | 0.0084 |
| 2.40 | 0.0082 | 0.0080 | 0.0078 | 0.0075 | 0.0073 | 0.0071 | 0.0069 | 0.0068 | 0.0066 | 0.0064 |
| 2.50 | 0.0062 | 0.0060 | 0.0059 | 0.0057 | 0.0055 | 0.0054 | 0.0052 | 0.0051 | 0.0049 | 0.0048 |
| 2.60 | 0.0047 | 0.0045 | 0.0044 | 0.0043 | 0.0041 | 0.0040 | 0.0039 | 0.0038 | 0.0037 | 0.0036 |
| 2.70 | 0.0035 | 0.0034 | 0.0033 | 0.0032 | 0.0031 | 0.0030 | 0.0029 | 0.0028 | 0.0027 | 0.0026 |
| 2.80 | 0.0026 | 0.0025 | 0.0024 | 0.0023 | 0.0023 | 0.0022 | 0.0021 | 0.0021 | 0.0020 | 0.0019 |
| 2.90 | 0.0019 | 0.0018 | 0.0018 | 0.0017 | 0.0016 | 0.0016 | 0.0015 | 0.0015 | 0.0014 | 0.0014 |
| 3.00 | 0.0013 | 0.0013 | 0.0013 | 0.0012 | 0.0012 | 0.0011 | 0.0011 | 0.0011 | 0.0010 | 0.0010 |
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