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 Home > CFA Exam Level 1 (12/2014) Study Notes > Study Session 3. Quantitative Methods: Application > Reading 10. Sampling and Estimation > LOS i. describe the properties of Student's t-distribution and calculate and interpret its degrees of freedom;
LOS i. describe the properties of Student's t-distribution and calculate and interpret its degrees of freedom;
Confidence intervals for the population mean.
Confidence intervals are typically constructed by using the following structure:

Confidence Interval = Point Estimate ± Reliability Factor x Standard Error

• Point estimate is the value of a sample statistic of the population parameter.
• Reliability factor is a number based on the sampling distribution of the point estimate and the degree of confidence (1 - α).
• Standard error refers to the standard error of the sample statistic that is used to produce the point estimate.
Whatever the distribution of the population, the sample mean is always the point estimate used to construct the confidence intervals for the population mean. The reliability factor and the standard error, however, may vary depending on three factors:
1. Distribution of population: normal or non-normal.
2. Population variance: known or unknown.
3. Sample size: large or small.
z-Statistic: a standard normal random variable

If a population is normally distributed with a known variance, z-statistic is used as the reliability factor to construct confidence intervals for the population mean.

In practice, the population standard deviation is rarely known. However, learning how to compute a confidence interval when the standard deviation is known is an excellent introduction to how to compute a confidence interval when the standard deviation has to be estimated.

Three values are used to construct a confidence interval for μ:

1. The sample mean (m);
2. The value of z (which depends on the level of confidence), and
3. The standard error of the mean (σ)m.
The confidence interval has m for its center and extends a distance equal to the product of z and in both directions. Therefore, the formula for a confidence interval is:

m - z σm <= μ <= m + z σm

For a (1 - α)% confidence interval for the population mean, the z-statistic to be used is zα/2. zα/2 denotes the points of the standard normal distribution such that α/2 of the probability falls in the right-hand tail.

Effectively, what is happening is that the (1 - α)% of the area that makes up the confidence interval falls in the center of the graph, that is, symmetrically around the mean. This leaves α% of the area in both tails, or α/2 % of area in each tail.

Commonly used reliability factors are as follows:

• 90% confidence intervals: z0.05 = 1.645. α is 10%, with 5% in each tail.
• 95% confidence intervals: z0.025 = 1.96. α is 5%, with 2.5% in each tail.
• 99% confidence intervals: z0.005 = 2.575. α is 1%, with 0.5% in each tail.
Example

Assume that the standard deviation of SAT verbal scores in a school system is known to be 100. A researcher wishes to estimate the mean SAT score and compute a 95% confidence interval from a random sample of 10 scores.

The 10 scores are: 320, 380, 400, 420, 500, 520, 600, 660, 720, and 780. Therefore, m = 530, N = 10, and σm= 100 / 101/2 = 31.62. The value of z for the 95% confidence interval is the number of standard deviations one must go from the mean (in both directions) to contain .95 of the scores.

It turns out that one must go 1.96 standard deviations from the mean in both directions to contain .95 of the scores. The value of 1.96 was found using a z table. Since each tail is to contain .025 of the scores, you find the value of z for which 1 - 0.025 = 0.975 of the scores are below. This value is 1.96.

All the components of the confidence interval are now known: m = 530, σm = 31.62, z = 1.96.
Lower limit = 530 - (1.96)(31.62) = 468.02
Upper limit = 530 + (1.96)(31.62) = 591.98

Therefore, 468.02 ≤ μ ≤ 591.98. This means that the experimenter can be 95% certain that the mean SAT in the school system is between 468 and 592. This also means if the experimenter repeatedly took samples from the population and calculated a number of different 95% confidence intervals using the sample information, on average 95% of those intervals would contain μ. Notice that this is a rather large range of scores. Naturally, if a larger sample size had been used, the range of scores would have been smaller.

The computation of the 99% confidence interval is exactly the same except that 2.58 rather than 1.96 is used for z. The 99% confidence interval is: 448.54 <= μ <= 611.46. As it must be, the 99% confidence interval is even wider than the 95% confidence interval.

Summary of Computations

1. Compute m = ∑X/N.
2. Compute σm = σ/N1/2
3. Find z (1.96 for 95% interval; 2.58 for 99% interval)
4. Lower limit = m - z σm
5. Upper limit = m + z σm
6. Lower limit <= μ <= Upper limit
Assumptions:
1. Normal distribution
2. σ is known
3. Scores are sampled randomly and are independent
There are three other points worth mentioning here:
1. The point estimate will always lie exactly at the midway mark of the confidence interval. This is because it is the "best" estimate for μ, and so the confidence interval expands out from it in both directions.
2. The higher the percentage of confidence, the wider the interval will be. This is because as the percentage is increased, a wider interval is needed to give us a greater chance of capturing the unknown population value within that interval.
3. The width of the confidence interval is always twice the part after the positive or negative sign, that is, twice the reliability factor x standard error. The width is simply the upper limit minus the lower limit.
It is very rare for a researcher wishing to estimate the mean of a population to already know its standard deviation. Therefore, the construction of a confidence interval almost always involves the estimation of both μ and σ.

Students' t-Distribution

When σ is known, the formula m - z σm <= μ <= m + z σm is used for a confidence interval. When σ is not known, σm = s/N1/2 (N is the sample size) is used as an estimate of σ and μ. Whenever the standard deviation is estimated, the t rather than the normal (z) distribution should be used. The values of t are larger than the values of z so confidence intervals when σ is estimated are wider than confidence intervals when σ is known. The formula for a confidence interval for μ when σ is estimated is:

m - t sm <= μ <= m + t sm

where m is the sample mean, sm is an estimate of σm, and t depends on the degrees of freedom and the level of confidence.

The t-distribution is a symmetrical probability distribution defined by a single parameter known as degrees of freedom (df). Each value for the number of degrees of freedom defines one distribution in this family of distributions. Like a standard normal distribution (e.g. a z-distribution), the t-distribution is symmetrical around its mean. Unlike a standard normal distribution, the t-distribution has the following unique characteristics.

• It is an estimated standardized normal distribution. When n gets larger, t approximates z (s approaches σ).
• The mean is 0, and the distribution is bell-shaped.
• There is not one t-distribution, but a family of t-distributions. All t-distributions have the same mean of 0. Standard deviations of these t-distributions differ according to the sample size, n.
• The shape depends on degrees of freedom (n - 1). The t-distribution is less peaked than a standard normal distribution, and has fatter tails (i.e. more probability in the tails).
• tα/2 tends to be greater than zα/2 for a given level of significance, α.
• Its variance is v/(v-2) (for v > 2), where v = n-1. It is always bigger than 1. As v increases, the variance approaches 1.

The value of t can be determined from a t table. The degrees of freedom for t is equal to the degrees of freedom for the estimate of σm which is equal to N-1.

A portion of t-table is presented as below:

Suppose the sample size (n) is 30, and the level of significance (α) is 5%. df = n - 1 = 29. tα/2 = t0.025 = 2.045 (Find the 29 df row, and then move to the 0.05 column).

Example

Assume a researcher is interested in estimating the mean reading speed (number of words per minute) of high-school graduates and computing the 95% confidence interval. A sample of 6 graduates was taken and the reading speeds were: 200, 240, 300, 410, 450, and 600. For these data,

• m = 366.6667
• sm = 60.9736
• df = 6-1 = 5
• t = 2.571
Therefore, the lower limit is: m - (t) (sm) = 209.904 and the upper limit is: m + (t) (sm) = 523.430. Therefore, the 95% confidence interval is:
209.904 <= μ <= 523.430

Thus, the researcher can be 95% sure that the mean reading speed of high-school graduates is between 209.904 and 523.430.

Summary of Computations

1. Compute m = ∑X/N.
2. Compute s
3. Compute σm = s/N1/2
4. Compute df = N-1
5. Find t for these df using a t table
6. Lower limit = m - t sm
7. Upper limit = m + t sm
8. Lower limit <= μ <= Upper limit
Assumptions:
1. Normal distribution
2. Scores are sampled randomly and are independent
Discuss the issues surrounding selection of the appropriate sample size

It's all starting to become a little confusing. Which distribution do you use?

When a large sample size (generally bigger than 30 samples) is used, a z table can always be used to construct the confidence interval. It does not matter if the population distribution is normal, or if the population variance is known or not. This is because the central limit theorem assures that when the sample is large, the distribution of the sample mean is approximately normal. However, the t-statistic is more conservative because the t-statistic tends to be greater than the z-statistic, and therefore using t-statistic will result in a wider confidence interval.

However, if there is only a small sample size, a t table has to be used to construct the confidence interval when the population distribution is normal and the population variance is not known.

If the population distribution is not normal, there is no way to construct a confidence interval from a small sample (even if the population variance is known).

Therefore, all else equal, you should try to select a sample larger than 30. The larger the sample size, the more precise the confidence interval.

In general, at least one of the following is needed:

• A normal distribution for the population.
• A sample size that is greater than or equal to 30.
If one or both of the above occur, then a z-table or t-table is used, dependent upon whether σ is known or unknown. If neither of the above occurs, then the question cannot be answered.

A summary of the situation is as follows:

• If the population is normally distributed, and the population variance is known, use a z-score irrespective of sample size.
• If the population is normally distributed, and the population variance is unknown, use a t-score irrespective of sample size.
• If the population is not normally distributed, and the population variance is known, use a z-score only if n >= 30, otherwise it cannot be done.
• If the population is not normally distributed, and the population variance is unknown, use a t-score only if n >= 30, otherwise it cannot be done.
 Practice Question 1When constructing confidence intervals for a normally distributed population, the t-distribution is used when: A. the population standard deviation is known. B. the sample size is less than 30, regardless of the population standard deviation. C. the population standard deviation is unknown and the sample size is less than 30.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 2You are thinking of using a t procedure to construct a 95% confidence interval for the mean of a population. You suspect the distribution of the population is not normal and may be skewed. Which of the following statement is correct? A. You should not use the t procedure because the population does not have a normal distribution. B. You may use the t procedure, provided your sample size is large, say at least 30. C. You may use the t- procedure because it is robust to non-normality.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 3What assumption is made when calculating a confidence interval around the mean for a sample with less than 30 data points? I. The sample is normally distributed. II. The population from which the sample was taken is normally distributed. III. The sample standard deviation is equal to the population standard deviation.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 4T-distributions are spread out _______(more or less) than a normal distribution with μ = 0, σ = 1.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 5A t-distribution with 30 d.f. is most similar to a _____ distribution. A. normal distribution with μ = 1 and σ2 = 1 B. normal distribution with μ = 0 and σ2 = 29 C. normal distribution with μ = 0 and σ2 = 1Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 6Crop researchers plant 100 plots with a new variety of corn. The average yield for these plots is x-bar = 130 bushels per acre. Assume that the yield per acre for the new variety of corn follows a normal distribution with unknown mean μ and standard deviation σ = 10 bushels per acre. A 90% confidence interval for μ is ______.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 7To assess the accuracy of a kitchen scale, a standard weight that is known to weigh 1 gram is repeatedly weighed a total of n times, and the mean x-bar of the weighing is computed. Suppose the scale readings are normally distributed with unknown mean μ and standard deviation σ = 0.01 g. How large should n be so that a 90% confidence interval for μ has a margin of error of 0.0001?Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 8For a random sample of size n = 100 from a population with s = 8, the error term, E, for a 90% confidence interval is ____(to nearest 0.01). A. 0.05 B. 1.32 C. 10Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 9If a random sample of size n = 100 from a population with s = 8, yields a sample mean of 56 then a 90% confidence interval for μ is ________. A. -1.32 < μ < 1.32 B. 54.98 < μ < 57.02 C. 54.68 < μ < 57.32Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 10To estimate the average number of miles tires at RoadStar will last, a random sample of 64 tires is tested. If the sample mean produced is 35,000 miles and the population standard deviation is 5000 miles then a 90% confidence interval is ______. A. 30,000 < m < 40,000 B. 26,775 < m < 43,225 C. 33,972 < m < 36,028Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 11To estimate the average length of their employee's telephone calls to within 0.1 minutes at a 90% confidence level, FoneJack, Inc. must randomly sample how many employee phone calls? [The population can be assumed normal with a standard deviation of 0.8 minutes.] A. 169 B. 14 C. 174Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 12For a 90% confidence interval for the population proportion, p, if the sample proportion, p', is 0.4 and the sample size is n = 100 then the error term, E, is ____(to nearest 0.001). A. 0.049 B. 0.004 C. 0.081Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 13In estimating the population proportion for the days in summer that have thunderstorms, p, if the sample proportion, p', is 0.35 and the sample size is n = 92 then the standard deviation of the sampling distribution is _____(to nearest 0.001). A. 0.650 B. 0.350 C. 0.050Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 14A local college wants to find the percentage of students with access to the Internet, at a 95% level of confidence how large of a sample would be needed to estimate the percentage to within 3 percentage points? A. 1068 B. 1849 C. 33Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 15A 95% confidence interval for a population mean will be ______ a 99% confidence interval for the same population mean. (Both calculations based on the same set of data.) A. longer than B. shorter than C. the same length as D. it depends on the particular set of dataCheck AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 16100 items are sampled from a normal distribution. The sample mean is 10 and the sample standard deviation is 2, then a 95% confidence interval for the sample mean must lie between: A. 9.608 and 10.392 B. 9.324 and 10.676 C. 9.108 and 10.892Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 17For the t-distribution, if a = 0.01 and n = 16 then the t-value is ______. A. 2.583 B. 2.602 C. 1.341Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 18For a normal population with s unknown, a sample of size n = 12 yields a sample mean of 54 and a sample standard deviation of 6 then at a 95% confidence level, the error term E is ____ (to nearest 0.01). A. 3.77 B. 3.39 C. 3.81Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 19For a normal population with s unknown, a sample of size n = 21 yields a sample mean of 32 and a sample standard deviation of 3 then at a 90% confidence interval for m is _____ (to nearest 0.01). A. 31.13 < m < 32.87 B. 30.35 < m < 33.65 C. 30.87 < m < 33.13Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 20As sample size increases, A. Critical values for t decrease. B. Degrees of freedom decrease. C. Confidence interval width increases.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 21True or False? If False, correct it. Confidence intervals can be shortened by increasing the sample size.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 22Select the correct statement(s). I. The main difference between the t and z statistic is that the t statistic has a mean of 1 and the z statistic has a mean of 0. II. The t-score is used when the sample size is 30 or more and you know the population standard deviation. III. Increasing the confidence level, while keeping other values constant, will keep the width of the interval unchanged.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 23A statistician is calculating the confidence interval for but the standard deviation of the sampled population is unknown. It is acceptable to use the sample standard deviation in place of the population standard deviation when I. when the sample distribution is normally distributed. II. the sample size is greater than 30. III. when the sample distribution behaves according to the Central Limit Theorem.A. I and II.B. I and III.C. I, II and III.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 24The main difference between the t and z statistic is that the t statisticA. is not normally distributed like the z statistic.B. is more variable than the z statistic.C. is less variable than the z statistic.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 25A statistician has randomly selected 25 samples from a production line for testing. In order for this sample to be a valid indication of the population, what assumption about the sample is necessary? I. The population variance is equal to the population mean. II. The population of paired differences must be normal. III. The population must have a normal distribution.A. III only.B. II and III.C. I and III.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 26When finding a confidence interval for the population mean, when is t-score used as the critical value (assume the population is normally distributed)?A. The sample size is less than 30 or you do not know the population standard deviation. B. The sample size is 30 or more or you know the population standard deviation.C. The sample size is less than 30 and you do not know the population standard deviation.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 27Which of the following is a true statement regarding the comparison of t-distributions to the standard normal distribution? I. The normal distribution is symmetrical whereas the t-distributions are slightly skewed. II. The proportion of area beyond a specific value of t is less than the proportion of area beyond the corresponding value of z. III. The greater the df, the more the t-distributions resemble the standard normal distribution.A. III only.B. I and III.C. II and III.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 28A client will move his investment account unless the portfolio manager earns at least a 10 percent rate of return on the account. The rate of return for the portfolio the portfolio manager has chosen has a normal probability distribution with an expected return of 19 percent and a standard deviation of 4.5 percent. What is the probability that the portfolio manager will keep this account?A. 0.950B. 1.000C. 0.975Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 29To estimate the average cost of a food-shopping event, Delcore, Inc. randomly sampled 100 shoppers and found a sample mean of \$72. Assuming a population standard deviation of \$5, a 99% confidence interval for average cost for a food-shopping event is _______.A. \$70.71 < m < \$73.29B. \$71.18 < m < \$72.82C. \$59.12 < m < \$84.88Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 30To estimate the average length of their employee's telephone calls, FoneJack, Inc. randomly sampled 25 employee phone calls. If the sample mean was 1.3 minutes and the population can be assumed normal with a standard deviation of 0.3 minutes then a 90% confidence interval for the phone calls is ______.A. 1.2 < m < 1.4B. 1.24 < m < 1.36C. 1.18 < m < 1.42Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 31To estimate the average weight of the red-crowned woodpecker to within 2 grams at a 95% confidence level, researchers must randomly sample how many red-crowned woodpeckers? [The population can be assumed normal with a standard deviation of 13 grams.]A. 282B. 70C. 163Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 32If the number of days in a 92-day summer period in which a thunderstorm occurred is 28 then a 90% confidence interval for the percentage of days in summer that have thunderstorms, p, is ______(to nearest 0.1%).A. 21.0% < p < 39.8%B. 25.6% < p < 35.2%C. 22.5% < p < 38.3%Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 33A researcher randomly samples 100 citizens of Florida and finds that 95 of the citizens are literate. The researcher then finds a 90% confidence interval. Which of the following is false?A. the researcher is 90% sure that the percentage of literate citizens is between 91.4% and 98.6%B. the sample proportion p' is 0.95 and the margin of error is 3.6%C. the confidence interval procedure is invalid because Np' and N(1 - p') are not both greater than 5Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 34A confidence interval was used to estimate the proportion of American new car owners who purchased domestic cars. A random sample of 58 new car owners generated the following 90% confidence interval: 0.376, 0.426. Based on the interval given, does the mean population proportion of new car owners who purchased domestic cars exceed 39%?A. Yes, and the researcher has 90% confidence in the result. B. The researcher cannot conclude that the mean exceeds 39% at the 90% confidence level. C. No, and the researcher has 90% confidence in the result.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 35Increasing the confidence level, while keeping other values constant, has what effect on the width of the confidence interval?A. There is not enough information to determine the effect. B. The error of estimate increases; therefore, the width of the interval increases. C. The error of estimate decreases; therefore, the width of the interval decreases.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 36A survey of 144 retail stores revealed that a particular brand and model of a VCR retails for \$375 with a standard deviation of \$20. If 95% and 98% confidence intervals are developed to estimate the true cost of the VCR, what difference would they have?A. Interval widthsB. Both interval widths and z-variatesC. Z-variatesCheck AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 37A hedge fund is interested in knowing how low an annual return their strategy might generate, as a once in a 100 year event. The strategy in question has an expected annual return of 20% with a standard deviation of 35%. You believe these returns are normally distributed. What is the low return that could be expected once in 100 years?A. -51.9%.B. -45.7%.C. -61.4%.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 38A retail client of yours is interested in knowing how high an annual return a major stock index might have, as a once in a twenty year event. The index in question has had an annual return of 11% with a standard deviation of 22%. You believe these returns have been normally distributed. What is the high return that could be expected once in twenty years?A. 54.1%.B. 22.0%.C. 47.2%.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 39A retail client of yours is interested in knowing how low an annual return a major stock index might have, as a once in a twenty year event. The index in question has had an annual return of 11% with a standard deviation of 22%. You believe these returns have been normally distributed. What is the low return that could be expected once in twenty years?A. -11.0%.B. -25.2%.C. -32.1%.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 40Which set of circumstances is most likely to result in a narrow confidence interval?A. large n and a degree of confidence of .95.B. large n and a degree of confidence of .99.C. small n and a degree of confidence of .95.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 41To estimate the average SAT scores for entering freshmen at universities, a random sample of 14 SAT scores is collected. If the sample mean produced a score of 1150 and the sample standard deviation is 150 points (σ is unknown) then a 95% confidence interval is _______.A. 1079.4 < μ < 1220.6B. 1063.4 < μ < 1236.6C. 1079.0 < μ < 1221.0Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 42To estimate the average length of their employee's telephone calls, FoneJack, Inc. randomly sampled 15 employee phone calls. If the sample mean was 1.3 minutes and the sample standard deviation of 0.3 minutes (s is unknown) then a 90% confidence interval for the phone calls is ______.A. 1.16 < m < 1.44B. 1.2 < m < 1.4C. 1.0 < m < 1.6Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 43There are 2,000 eligible voters in a precinct. Despite protests from knowledgeable persons that a sample size of 500 was too large in relation to the total, the 500 selected at random were asked to indicate whether they planned to vote for the Democratic incumbent or the Republican challenger. Of the 500 surveyed, 350 said they were going to vote for the Democratic incumbent. Using the 0.99 confidence coefficient, what are the confidence limits for the proportion who plan to vote for the Democratic incumbent?A. 0.397 and 0.797B. 0.612 and 0.712C. None of these answersCheck AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 44Calculate an 80% confidence interval for a population mean. You have a sample of 21, a sample mean of -25%, and a sample standard deviation of 10%. The sample appears to be approximately normally distributed.A. [-26%, -24%].B. [-28%, -22%].C. [-27%, -23%].Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 45Calculate a 90% confidence interval for a population mean. You have a sample of 21, a sample mean of -25%, and a sample standard deviation of 10%. The sample appears to be approximately normally distributed.A. [-28.9%, -21.1%].B. [-27.9%, -22.1%].C. [-28.1%, -21.9%].Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 46If the size of the sample being used is increased, then the width of a 0.95 confidence interval estimate for a population mean will:A. Become narrower.B. Become wider.C. Not be changed.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 47Which statement(s) regarding the selection of an appropriate sample size is/are false?I. If the population is normally distributed, and the population variance is known, use a t-score irrespective of sample size.II. If the population is normally distributed, and the population variance is unknown, use a z-score irrespective of sample size.III. If the population is not normally distributed, and the population variance is known, use a F-score only if n >= 30, otherwise it cannot be done. IV. If the population is not normally distributed, and the population variance is unknown, use a X2-score only if n >= 30, otherwise it cannot be doneA. I and II.B. II, III and IV.C. I, II, III and IV.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 48Consider the following statements: I. The t distribution is a symmetrical probability distribution defined by a single parameter known as the degrees of freedom. II. When sample size is small and population variance is not known, the t distribution is used. III. The use of a t reliability factor is appropriate when the population variance is unknown but we have a large sample and can rely on the central limit theorem to ensure approximate normality of the distribution of the sample mean. Which is true?A. I and II.B. I and III.C. I, II and III.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 49Consider the following statements relating to sample size and confidence intervals:I. The width of a confidence interval decreases as the sample size increases.II. When width of a confidence interval decreases this is due to standard error becoming smaller.III. More precise results are obtained when the sample size is increased.Which is true?A. II and III.B. I and III.C. I, II and III.Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 50A normal distribution has a mean of 25% and a standard deviation of 33%. What is the 95% confidence interval around the mean, i.e., if 95% of observations can be expected to fall in this range?A. -29.29, 79.29B. -8.00, 58.00C. -39.68, 89.68Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 51The mean of sample with a known population variance of 25 was estimated as 12.5. The sample size was 22 and the sample variance was 28. Which of the following represents the appropriate 95% confidence interval for the normally distributed population?A. 10.41, 14.59B. 10.29, 14.71C. 10.74, 14.26Check AnalystNotes for the correct answer and a detailed explanation.
 Practice Question 52An increase in which of the following items is most likely to result in a wider confidence interval for the population mean?A. Sample sizeB. Degrees of freedomC. Reliability factorCheck AnalystNotes for the correct answer and a detailed explanation.
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