2.4 Sample Moments and Hypothesis Tests

2.4 Sample Moments and Hypothesis Tests

Question 1

Which of the following statements regarding hypothesis testing is incorrect?

A. Type II error refers to the failure to reject the null hypothesis when it is actually false.
B. Hypothesis testing is used to make inferences about the parameters of a given population on the basis of statistics computed form a sample that is drawn from that population.
C. All else being equal, the decrease in the chance of making a Type I error comes at the cost of increasing the probability of making a Type II error.
D. The $\text{p-value}$ decision rule is to reject the null hypothesis if the p-value is greater than the significance level.

Answer: D
We would reject the null hypothesis if the observed $\text{p-value}$ is lower than the significance level.

• The p-value is the smallest level of significance at which the null hypothesis can be reject.
• If $\text{p-value}<\alpha$, then reject the null hypothesis
• When $\text{p-value}$ decreases, it is easier to reject the null hypothesis.
• The way to reduce both errors is to increase the sample size.

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Question 2

An investment analyst takes a random sample of $100$ aggressive equity funds and calculates the average beta as $1.7$. The sample betas have a standard deviation of $0.4$. Using a $95\%$ confidence interval and a z-statistic, which of the following statements about the confidence interval and its interpretation is most likely accurate? The analyst can be confident at the $95\%$ level that the interval:

A. $0.916$ to $2.484$ includes the mean of the sample betas.
B. $1.622$ to $1.778$ includes the mean of the sample betas.
C. $0.916$ to $2.484$ includes the mean of the population beta.
D. $1.622$ to $1.778$ includes the mean of the population beta.

Answer: D
$CI=\overline{x}\pm z_{\alpha /2}\frac{\sigma }{\sqrt{n}}=1.7\pm 1.96\times \frac{0.4}{\sqrt{100}}$

For normal distribution

• $68\%$ confidence interval is $\mu \pm \sigma$
• $90\%$ confidence interval is $\mu \pm 1.645\sigma$
• $95\%$ confidence interval is $\mu \pm 1.96\sigma$
• $99\%$ confidence interval is $\mu \pm 2.58\sigma$

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Question 3

The annual returns for a portfolio are normally distributed with an expected value of £$50$ million and a standard deviation of £$25$ million. Which of the following amounts is closest to the probability that the value of the portfolio one year from today will be between £$91.13$ million and £$108.25$ million?

A. $0.025$
B. $0.040$
C. $0.075$
D. $0.090$

Answer: B
Calculate the standardized variable corresponding to the outcomes:
$Z_1=(91.13-50)/25=1.645$, $Z_2=(108.25-50)/25=2.33$

The cumulative normal distribution gives cumulative probabilities of:
$P(Z_1)=0.95$, $P(Z_2)=0.99$

The probability that the outcome will lie between $Z_1$ and $Z_2$ is the difference: $0.99-0.95=0.04$

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Question 4

An analyst is concerned that the trading strategy she recently identified has generated a statistically insignificant result and has asked for guidance in assessing the strategy. A result is statistically significant if it is:

A. Unlikely to have occurred merely by chance, and the p-value$\text{p-value}$ is less than the significance level.
B. Likely to have occurred merely by chance, and the $\text{p-value}$ is less than the significance level.
C. Unlikely to have occurred merely by chance, and the $\text{p-value}$ is greater than the significance level.
D. Likely to have occurred merely by chance, and the $\text{p-value}$ is greater than the significance level.

Answer: A
A result is statistically significant if it is unlikely to have happened by chance. The decision rule is to reject the null hypothesis if the$\text{p-value}$ is less than the significance level. If the $\text{p-value}$ is less than the significance level, then we conclude that the sample estimate is statistically different than the hypothesized value.

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Question 5

Which of the following statements are NOT true?

I Type I error occurs when the null hypothesis is not rejected when it is actually false.
II Type II error occurs when the null hypothesis is rejected when it is actually true.
III Type I error occurs when the alternate hypothesis is wrongly accepted.
IV Minimizing the probability of Type II error maximizes the power of the test.

A. I and II
B. I and III
C. II and IV
D. I, II and IV

Answer: A
In hypothesis testing we accept the alternate hypothesis if the null hypothesis has been rejected. Type I error happens if the null hypothesis is rejected when it is actually true. Type II error happens if the null hypothesis is accepted when it is actually false. The power of the test is the probability of correctly rejecting the null hypothesis (when it is false), so minimizing Type II errors would maximize the power of the test.

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Question 6

Adam Farman has been asked to estimate the volatility of a technology stock index.He has identified a statistic which has an expected value equal to the population volatility and has determined that increasing his sample size will decrease the sampling error for this statistic. His statistic can best be described as:

A. unbiased and efficient
B. unbiased and consistent
C. efficient and consistent
D. unbiased only

Answer: B
An unbiased estimator has an expected value equal to the true value of the population parameter.
A consistent estimator is more accurate the greater the sample size.
An efficient estimator has the sampling distribution that is less than that of any other unbiased estimator.

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Question 7

If the mean P/E of $30$ stocks in a certain industrial sector is $18$ and the sample standard deviation is $3.5$, standard error of the mean is CLOSEST to:

A. $0.12$
B. $0.34$
C. $0.64$
D. $1.56$

Answer: C
The standard error of the sample mean is the standard deviation of the distribution of the sample means. And it is calculated as:
$SE=\sigma /\sqrt{n}$,where $\sigma$, the population standard deviation is known.
$SE=s/\sqrt{n}$, where $S$, is the sample standard deviation.
So, standard error of the mean, $3.5/\sqrt{30}=0.64$.

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Question 8

PE2018Q68 / PE2019Q68 / PE2020Q68 / PE2021Q68 / PE2022PSQ15 / PE2022Q68
A risk manager is calculating the $\text{VaR}$ of a fund with a data set of $25$ weekly returns. The mean and standard deviation of weekly returns are $7\%$ and $15\%$, respectively. Assuming that weekly returns are independent and identically distributed, what is the standard deviation of the mean of the weekly returns?

A. $0.4\%$
B. $0.7\%$
C. $3.0\%$
D. $10.0\%$

Answer: C
Learning Objective: Calculate and interpret the sample mean and sample variance

In order to calculate the standard deviation of the mean of weekly returns, we must divide the standard deviation of the weekly returns by the square root of the sample size. Therefore the correct answer is $15\%/\sqrt{25}=3\%$.

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Question 9

If the variance of the sampling distribution of an estimator is smaller than all other unbiased estimators of the parameter of interest, the estimator is:

A. Reliable
B. Efficient
C. Unbiased
D. Consistent

Answer: B
If the probability distribution of an estimator has an expected value equal to the parameter it is supposed to be estimating, it is said to be unbiased.

Between two candidate estimators, the one with a smaller variance is said to use the information in the data more efficiently.

When the probability that an estimator is within a small interval of the true value approaches $1$, it is said to be a consistent estimator.

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Question 10

Analyst Rob has identified an estimator, denoted $T(.)$, which qualifies as the bes,tlinear unbiased estimator (BLUE). If $T(.)$ is BLUE, which of the following must also necessarily be TRUE?

A. $T(.)$ must have the minimum variance among all possible estimators.
B. $T(.)$ must be the most efficient (the “best”) among all possible estimators
C. It is possible that $T(.)$ is the maximum likelihood estimator(MLE) of variance; i.e., $\sum ([X-E(X){]}^2)/(n-1)$
D. Among the class of unbiased estimators that are linear, $T(.)$ has the smallest variance.

Answer: D
The mean estimator is the Best Linear Unbiased Estimator (BLUE) of the population mean when the data are iid.

• Linear: Linear estimators of the mean can be expressed as: $\hat{\mu }={\sum }_{i-1}^n{w}_i{X}_i$
• Best: the smallest variance among all linear and unbiased estimators.
• It does not, however, imply that there are no superior estimators to the sample mean. It only implies that these estimators must either be biased or nonlinear.

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Question 11

PE2018Q64 / PE2019Q64 / PE2020Q64 / PE2021Q64 / PE2022Q64
For a sample of the past $30$ monthly stock returns for McCreary, Inc., the mean return is $4\%$ and the sample standard deviation is $20\%$. Since the population variance is unknown, the standard error of the sample is estimated to be:

$$S_X=\frac{20\%}{\sqrt{30}}=3.65\%$$

The related t-table values are ($t_{ij}$ denotes the $(100-j{)}^{th}$ percentile of t-distribution value with $i$ degrees of freedom):

What is the $95\%$ confidence interval for the mean monthly return?

A. $[-3.464\%,11.464\%]$
B. $[-3.453\%,11.453\%]$
C. $[-2.201\%,10.201\%]$
D. $[-2.194\%,10.194\%]$

Answer: A
Learning Objective: Construct and apply confidence intervals for one-sided and two-sided hypothesis tests, and interpret the results of hypothesis tests with a specific confidence level.

Here the t-reliability factor is used since the population variance is unknown. Since there are $30$ observations, the degrees of freedom are $30-1=29$.

The t-test is a two-tailed test. So the correct critical t-value is $t_{29,2.5}=2.045$, thus the $95\%$ confidence interval for the mean return is:
$[4\%-2.045\times \frac{20\%}{\sqrt{30}},4\%+2.045\times \frac{20\%}{\sqrt{30}}]=[-3.464\%,11.464\%]$

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Question 12

PE2018Q43 / PE2019Q43 / PE2020Q43 / PE2021Q43 / PE2022Q43
Using the prior $12$ monthly returns, an analyst estimates the mean monthly return of stock XYZ to be $-0.75\%$ with a standard error of $2.70\%$.

ONE-TAILED T-DISTRIBUTION TABLE

Using the t-table above, the $95\%$ confidence interval for the mean return is between:

A. $-6.69\%$ and $5.19\%$
B. $-6.63\%$ and $5.15\%$
C. $-5.60\%$ and $4.10\%$
D. $-5.56\%$ and $4.06\%$

Answer: A
Learning Objective: Construct and apply confidence intervals for one-sided and two-sided hypothesis tests and interpret the results of hypothesis tests with a specific level of confidence.

The confidence interval is equal to the mean monthly return plus or minus the t-statistic times the standard error.

To get the proper t-statistic, the $0.025$ column must be used since this is a two-tailed interval.
Since the mean return is being estimated using the sample observations, the appropriate degrees of freedom to use is equal to the number of sample observations minus $1$.
Therefore we must use $11$ degrees of freedom and therefore the proper statistic to use from the t-distribution is $2.201$. The proper confidence interval is: $-0.75\%\pm (2.201\times 2.70\%)$ or $-6.69\%$ to $5.19\%$.

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Question 13

A quantitative analyst used a simulation to forecast the S&P 500 index value at the end of the year with an index value of $1.800$ at the beginning of the year. He generated $200$ scenarios and calculated the average index value at year-end to be $1.980$, with a $95\%$ confidence interval of $(1.940,2.020)$. In order to improve the accuracy of the forecast, the quantitative analyst increased the number of scenarios to attain a new $95\%$ confidence interval of $(1.970,1.9901)$ with the same sample mean and the same sample standard deviation. How many scenarios were used to generate this result?

A. $400$
B. $800$
C. $1600$
D. $3200$

Answer: D
Confidence interval estimate $\hat{x}\pm Z_{\alpha /2}\frac{\sigma }{\sqrt{n}}$

When there are 200 scenarios, $(2.020-1.940)/2=1.96\frac{\sigma }{\sqrt{2000}}\to \sigma =0.2886$

Then $(1.9901-1.97)/2=1.96\frac{0.2886}{\sqrt{n}}\to n\approx 3200$

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Question 14

When testing a hypothesis, which of the following statements is correct when the level of significance of the test is decreased?

A. The likelihood of rejecting the null hypothesis when it is true decreases.
B. The likelihood of making a Type I error increases.
C. The null hypothesis is rejected more frequently, even when it is actually false.
D. The likelihood of making a Type Il error decreases.

Answer: A
Significance level = P(Type I)

Decreasing the level of significance of the test decreases the probability of making a Type I error and hence makes it more difficult to reject the null when it is true. However, the decrease in the chance of making a Type I error comes at the cost of increasing the probability of making a Type Il error, because the null is rejected less frequently, even when it is actually false.

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Question 15

An oil industry analyst with a large international bank has constructed a sample of $1000$ individual firms on which she plans to perform statistical analyses. She considers either decreasing the level of significance used to test hypotheses from $5\%$ to $1\%$, or removing $500$ state-run firms from her sample. What impact will these changes have on the probability of making Type I and Type Il errors

Answer: B
Type I error is the error of rejecting a hypothesis when it is true. Type Il error is the error of accepting a false hypothesis.

A decrease in the level of significance decreases P(Type I error) but increases P(Type Il error). Reducing the sample size increases P(Type Il error).

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Question 16

According to the Basel back-testing framework guidelines, penalties start to apply if there are five or more exceptions during the previous year. The Type I error rate of this test is $11$ percent. If the true coverage is $97$ percent of exceptions instead of the required $99$ percent, the power of the test is $87$ percent. This implies that there is a (an):

A. $89\%$ probability regulators will reject the correct model.
B. $11\%$ probability regulators will reject the incorrect model.
C. $87\%$ probability regulators will not reject the correct model.
D. $13\%$ probability regulators will not reject the incorrect model.

Answer: D
The power of the test refers to the probability of rejecting an incorrect model, which is one minus the probability of not rejecting an incorrect model. Given that the power of the test is 87 percent, the probability of a type II error, the probability of not rejecting the incorrect model is $1-0.87=13\%$.

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Question 17

Bob tests the null hypothesis that the population mean is less than or equal to $45$. From a population size of $3,000,000$ people, $81$ observations are randomly sampled. The corresponding sample mean is $46.3$ and sample standard deviation is $4.5$. What is the value of the appropriate test statistic for the test of the population mean, and what is the correct decision at the $1$ percent significance level?

A. $z=0.29$, and fail to reject the null hypothesis.
B. $z=2.60$, and reject the null hypothesis.
C. $t=0.29$, and accept the null hypothesis.
D. $t=2.60$, and neither reject nor fail to reject the null hypothesis.

Answer: B
A is incorrect. The denominator of the z-test statistic is standard error instead of standard deviation. If the denominator takes the value of standard deviation $4.5$, instead of standard error $4.5/\sqrt{81}$, the z-test statistic computed will be $z=0.29$, which is incorrect.

B is correct. The population variance is unknown but the sample size is large ($>30$). The test statistics is: $z=(46.3-45)/(4.5/\sqrt{81})=2.60$. Decision rule: reject $H_0$ if z(computed)>z(critical). Therefore, reject the null hypothesis because the computed test statistics of $2.6$ exceeds thecritical z-value of $2.33$.

C and D are incorrect because z-test (instead of t-test) should be used for sample size $81\ge 30$.

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Question 18

Which one of the following four statements about hypothesis testing holds true if the level of significance decreases from $5\%$ to $1\%$?

A. It becomes more difficult to reject a null hypothesis when it is actually true.
B. The probability of making a Type I error increases.
C. The probability of making a Type Il error decreases.
D. The failure to reject the null hypothesis when it is actually false decreases to $1\%$.

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Question 19

Which of the following statements regarding hypothesis testing is correct?

A. Type Il error refers to the failure to reject the $H1$ when it is actually false.
B. Hypothesis testing is used to make inferences about the parameters of a given population on the basis of statistics computed for a sample that is drawn from another population.
C. All else being equal, the decrease in the chance of making a Type I error comes at the cost of increasing the probability of making a Type Il error.
D. If the $\text{p-value}$ is greater than the significance level, then the statistics falls into the reject intervals.

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Question 20

An analyst wants to test whether the standard deviation of return from pharmaceutical stocks is lower than $0.2$. For this purpose, he obtains the following data from a sample of 30 pharmaceutical stocks. Mean return from pharmaceutical stocks is $8\%$. Standard deviation of return from pharmaceutical stocks is $12\%$. Mean return from the market is $12\%$. Standard deviation of return from the market is $6\%$. What is the appropriate test statistic for this test?

A. t-statistic
B. z-statistic
C. F-statistic
D. ${\chi }^2$ statistic

Answer: D
Tests of the variance (or standard deviation) of a population requires the chi-squared test.

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Question 21 (PE2018Q30)

A risk manager is examining a Hong Kong trader’s profit and loss record for the last week, as shown in the table below:

The profits and losses are normally distributed with a mean of $4.5$ million HKD and assume that transaction costs can be ignored. Part of the t-table is provided below:

Percentage Point of the t-Distribution $P(T>t)=\alpha$

According to the information provided above, what is the probability that this trader will record a profit of at least HKD $30$ million on the first trading day of next week?

A. About $15\%$
B. About $20\%$
C. About $80\%$
D. About $85\%$

Answer: B
Learning Objective: Apply and interpret the t-statistic when the sample size is small.

When the population mean and population variance are not known, the t-statistic can be used to analyze the distribution of the sample mean.

Sample mean: $(10+80+90-60+30)/5=30$
Unbiased sample variance: $[(-20{)}^2+50^2+60^2+(-90{)}^2]/4=14600/4=3650$
Sample standard error: $\sqrt{3650/5}=27.0185$
Population mean of return distribution: $4.5$ million HKD
Therefore the t-statistic: $(30-4.5)/27.02=0.9438$

Because we are using the sample mean in the analysis, we must remove 1 degree of freedom before consulting the t-table; therefore 4 degrees of freedom are used. According to the table, the closest possibility is $20\%$.

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Question 22

Hedge Fund has been in existence for two years. Its average monthly return has been $6\%$ with a standard deviation of $5\%$. Hedge Fund has a stated objective of controlling volatility as measured by the standard deviation of monthly returns. You are asked to test the null hypothesis that the volatility of Hedge Fund’s monthly returns is equal to $4\%$ versus the alternative hypothesis that the volatility is greater than $4\%$. Assuming that all monthly returns are independently and identically normally distributed, and using the tables below, what is the correct test to be used and what is the correct conclusion at the $2.5\%$ level of significance?

t Table: Inverse of the one-tailed probability of the Student's t-distribution Chi-Square Table: Inverse of the one-tailed probability of the Chi-Squared distribution

A. t-test; reject the null hypothesis
B. Chi-square test; reject the null hypothesis
C. t-test; do not reject the null hypothesis
D. Chi-square test; do not reject the null hypothesis

Answer: D
Null Hypothesis: ${\sigma }^2=4{\%}^2=0.0016$

Alternative Hypothesis: ${\sigma }^2>0.0016$

Since $\frac{(24-1)\times 0.05^2}{0.04^2}\approx 36<{\chi }_{2.5,23}^2=38.0757$, you would not reject the null hypothesis.

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Question 23

Using a sample size of $61$ observations, an analyst determines that the standard deviation of the returns from a stock is $21\%$. Using a $0.05$ significance level, the analyst:

A. Can conclude that the standard deviation of returns is higher than $14\%$.
B. Cannot conclude that the standard deviation of returns is higher than $14\%$.
C. Can conclude that the standard deviation of returns is not higher than $14\%$.
D. None of the above.

Answer: A
The required test for testing the variance is the chi-squared test.${\chi }^2=(61-1)\times (\frac{21\%}{14\%}{)}^2=135$

To test whether the standard deviation is higher ($H_0$: standard deviation is lower than or equal to $14\%$), the critical value of chi-squared will be $79.08$ (using $\text{df}=60$ and $p=0.05$).

Since the test statistic is higher than the critical value, the analyst can reject the null hypothesis and concludes that the standard deviation of returns is higher than $14\%$.

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Question 24

Based on $21$ daily returns of an asset, a risk manager estimates the standard deviation of the asset’s daily returns to be $2\%$. Assuming that returns are normally distributed and that there are $260$ trading days in a year, what is the appropriate Chi-square test statistic if the risk manager wants to test the null hypothesis that the true annual volatility is $25\%$ at a $5\%$ significance level?

A. $25.80$
B. $33.28$
C. $34.94$
D. $54.74$

Answer: B
The formula for the Chi-squared test statistic is: $\frac{(n-1)S^2}{{\sigma }^2}$

Since we are given a daily standard deviation, we must first annualize it by multiplying it by the square root of the number of trading days.

Therefore: $S=2\%\times \sqrt{260}=0.3225$

And the Chi-squared test statistic is: $(21-1)\times 0.3225^2/0.25^2=33.28$

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Question 25

Assuming the height of a tree in a forest follows a normal distribution, there are more than $10,000$ trees. The sample size of test is $200$, the $95\%$ confidence interval of the sample mean of the height is $11$ to $35$ meters based on a z-statistic, the standard error of mean height is closest to:

A. $1.96$
B. $3.58$
C. $6.12$
D. $10.50$

Answer: C
$11=\mu -1.96\times SE$, $35=\mu +1.96\times SE$, so $SE=6.12$