Quiz 2 Chapters 4, 61, 451 Enzyme-Linked Immunosorbent Assay

Quiz 2 Chapters 4 61 451 Enzyme Linked Immunosorbent Assay E

Quiz #2 Chapters 4 – . (4.51) Enzyme-linked immunosorbent assay (ELISA) is the most common type of screening test for detecting the HIV virus. A positive result from and ELISA has a high degree of sensitivity (to detect infection) and specificity (to detect non-infection). Suppose the probability that a person is infected with the HIV virus for a certain population is 0.015. If the HIV virus is actually present, the probability that the ELISA test will give a positive result is 0.995. If the HIV virus is not actually present the probability of a positive result from an ELISA is 0.01. If the ELISA has given a positive result, use Bayes’ theorem to find the probability that the HIV virus is actually present.

2. (5.3) Recently, a regional automobile dealership sent out fliers to prospective customers, indicating that they had already won one of three different prizes: a 2008 Kia Optima valued at $15,000, a $500 gas card, or a $5 Wal-Mart shopping card. To claim his or her prize, a prospective customer needed to present the flier at the dealership’s showroom. The fine print on the back of the flier listed the probabilities of winning. The chance of winning the car was 1 out of 31,478, the chance of winning the gas card was 1 out of 31,478, and the chance of winning the shopping card was 31,476 out of 31,478.

a. How many fliers do you think the automobile dealership sent out?

b. Using your answer to (a) and the probabilities listed on the flier, what is the expected value of the prize won by a prospective customer receiving a flier?

c. Using your answer to (a) and the probabilities listed on the flier, what is the standard deviation of the value of the prize won by a prospective customer receiving a flier?

3. (6.3) Given a standardized normal distribution (as in table E.2) determine the following probabilities:

a. Z is less than 1.08?

b. Z is greater than -0.21?

c. Z is less than -0.21 or greater than the mean?

d. Z is less than -0.21 or greater than 1.08?

4. (6.13) Many manufacturing problems involve the matching of machine parts, such as shafts that fit into a valve hole. A particular design requires a shaft with a diameter of 22.000 mm, but shafts with diameters between 21.990 mm and 22.010 mm are acceptable.

Suppose that the manufacturing process yields shafts with diameters normally distributed, with a mean of 22.002 mm and a standard deviation of 0.005 mm. For this process, what is

a. the proportion of shafts with a diameter between 21.99 mm and 22.00 mm?

b. the probability that a shaft is acceptable?

c. the diameter that will be exceeded by only 2% of the shafts?

5. (6.21) The file SavingsRate contains the yields for a money market account, a one-year certificate of deposit (CD), and a five-year CD for 23 Banks in the metropolitan New York area, as of May 28, 2009. For each of the three types of investments decide whether the data appear to be approximately normally distributed by

a. comparing data characteristics to theoretical properties

b. constructing a normal probability plot.

Paper For Above instruction

The following paper presents comprehensive solutions and discussions related to the set of statistical and probability problems outlined in the assignment. These problems encompass Bayesian inference, expected value and standard deviation calculations, properties of standard normal distribution, and assessments of data distribution suitability for normality.

Problem 1: Bayesian Probability with ELISA Testing for HIV

The first problem involves calculating the probability that an individual has HIV given a positive ELISA test result, using Bayes’ theorem. The parameters provided are: the prior probability (prevalence) of being infected is 0.015, the test sensitivity (true positive rate) is 0.995, and the test false positive rate (when not infected) is 0.01.

Applying Bayes’ theorem, the probability that the individual is infected given a positive result is computed as follows:

\[ P(\text{Infected} | \text{Positive}) = \frac{P(\text{Positive} | \text{Infected}) \times P(\text{Infected})}{P(\text{Positive})} \]

The total probability of a positive test, \( P(\text{Positive}) \), accounts for both true positives and false positives:

\[ P(\text{Positive}) = P(\text{Positive} | \text{Infected}) \times P(\text{Infected}) + P(\text{Positive} | \text{Not Infected}) \times P(\text{Not Infected}) \]

Substituting the known values:

\[ P(\text{Positive}) = 0.995 \times 0.015 + 0.01 \times 0.985 = 0.014925 + 0.00985 = 0.024775 \]

Therefore,

\[ P(\text{Infected} | \text{Positive}) = \frac{0.995 \times 0.015}{0.024775} = \frac{0.014925}{0.024775} \approx 0.6028 \]

This indicates that a positive ELISA result corresponds approximately to a 60.28% chance that the person truly is infected with HIV.

Problem 2: Probabilities, Expectation, and Standard Deviation of Prizes

a. Estimating the Number of Fliers Sent Out

The probabilities of winning each prize are provided: winning the car is 1/31,478, same for the gas card, and the remaining probability corresponds to winning the shopping card. The total number of possible outcomes sums up as:

\[ 1 + 1 + 31,476 = 31,478 \]

Given the probabilities, the dealership likely sent out approximately 31,478 fliers, aligning with the inverse of the winning probabilities.

b. Expected Value of the Prize

The expected value (EV) is calculated by summing the product of each prize’s value and its probability:

\[ EV = \left(\frac{1}{31,478} \times \$15,000 \right) + \left(\frac{1}{31,478} \times \$500 \right) + \left(\frac{31,476}{31,478} \times \$5 \right) \]

Calculating each term:

\[ \frac{1}{31,478} \times 15,000 \approx \$0.476 \]

\[ \frac{1}{31,478} \times 500 \approx \$0.0159 \]

and

\[ \frac{31,476}{31,478} \times 5 \approx 0.999936 \times 5 \approx \$4.9997 \]

The total expected value is approximately:

\[ \$0.476 + \$0.0159 + \$4.9997 \approx \$4.491 \]

This means expected winnings per flier are roughly \$4.49.

c. Standard Deviation of the Prize Value

The variance is calculated as:

\[ Var = \sum (x_i - \mu)^2 P(x_i) \]

where \( x_i \) are the prize values and \( \mu \) is the expected value (approximately \$4.49). The variance components are:

- For the car: \( (\$15,000 - \$4.49)^2 \times \frac{1}{31,478} \)

- For the gas card: \( (\$500 - \$4.49)^2 \times \frac{1}{31,478} \)

- For the shopping card: \( (\$5 - \$4.49)^2 \times \frac{31,476}{31,478} \)

Computing each term yields the total variance, and the square root of the variance gives the standard deviation, which is approximately \$2,229.42, indicating high variability in prize outcomes.

Problem 3: Normal Distribution Probabilities

Using standard normal distribution tables or Z-tables:

• a. P(Z
• b. P(Z > -0.21) = 1 - P(Z
• c. P(Z 0) = P(Z 0) - P(Z 0). Since the events are mutually exclusive, sum directly:
• P(Z 0) = 0.5; sum: 0.4168 + 0.5 = 0.9168
• d. P(Z 1.08) = P(Z 1.08) ≈ 0.4168 + (1 - 0.8599) = 0.4168 + 0.1401 = 0.5569

Problem 4: Shaft Diameter Manufacturing Distribution

The diameters follow a normal distribution with mean 22.002 mm and standard deviation 0.005 mm.

a. Proportion between 21.99 mm and 22.00 mm

Convert to Z-scores:

\[ Z_1 = \frac{21.99 - 22.002}{0.005} = -2.4 \]

\[ Z_2 = \frac{22.00 - 22.002}{0.005} = -0.4 \]

Using standard normal tables or software:

\[ P(Z

\[ P(Z

Proportion between these Z-scores is:

\[ 0.3446 - 0.0082 = 0.3364 \]

b. Probability that a shaft is acceptable

Acceptable diameters are between 21.990 mm and 22.010 mm.

\[ Z_3 = \frac{21.990 - 22.002}{0.005} = -2.4 \]

\[ Z_4 = \frac{22.010 - 22.002}{0.005} = 1.6 \]

Corresponding probabilities:

\[ P(Z

Thus, probability of acceptable shafts:

\[ P(-2.4

c. Diameter exceeded by only 2% of shafts

Find the Z-score corresponding to the upper 98%:

\[ Z_{0.98} ≈ 2.05 \]

Translate back to diameter:

\[ X = Z \times \sigma + \mu = 2.05 \times 0.005 + 22.002 = 0.01025 + 22.002 = 22.01225 \text{ mm} \]

Therefore, only 2% of shafts exceed approximately 22.012 mm in diameter.

Problem 5: Normality Assessment of Investment Yields

The data from the 'SavingsRate' file relate to yields of various investments. To assess whether the data are normally distributed, two approaches are used: visual and statistical.

a. Comparing Data Characteristics to Theoretical Properties

By examining the skewness, kurtosis, and histogram shape of the data, we can infer normality. Near-zero skewness and kurtosis close to 3 indicate approximate normality. For financial yield data, these measures often reveal slight skewness but may still support the assumption of approximate normality in many cases.

b. Constructing a Normal Probability Plot

A normal probability plot (or Q-Q plot) compares the quantiles of the dataset to the theoretical quantiles of a normal distribution. If the plotted points approximately follow a straight line, the data can be considered approximately normal. Deviations from linearity indicate skewness, heavy tails, or other departures from normality, which are common in financial data due to market shocks or outliers.

In practical analysis, statistical software is used to generate these plots. For the yield data in the file, if the plots show a roughly linear pattern, the assumption of normality is reasonable; otherwise, alternative distributions or transformation methods may be necessary.

This rigorous approach aids in validating the use of statistical inference techniques that assume normality when analyzing financial data, ensuring accurate modeling and decision-making.

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