1. The table shows the mass of fibres in 100 grams of each food item below.

1. A small tin of beans contains 200 grams. If you are going to eat two-thirds (2/3) of this small tin of beans, how much fibres will you consume? Show your calculations.
2. List TWO functions of fibres in the diet.
3. What is the consequence of lacking in fibres in the diet? (1 mark)
4. Potatoes contain carbohydrates. Which form of carbohydrate do they contain? (1 mark)
5. Briefly describe the digestion of this form of carbohydrate described in (iv) in mouth and small intestine before absorption.

Given data from a completely randomized design experiment:

Treatment 1 = {3.8, 1.2, 4.1, 5.5, 2.3}

Treatment 2 = {5.4, 2.0, 4.8, 3.8}

Treatment 3 = {1.3, 0.7, 2.2}

a.) Calculate the treatment means and variances for each of the 3 treatments above.

b.) Use statistical software to complete the ANOVA table.

c.) In words, what is the null and alternative hypotheses for the ANOVA F-test?

d.) Test the null hypothesis that µ1=µ2=µ3against the alternative hypothesis that at least two means differ. Use α = .01.

e.) Explain in words what the ANOVA test tells us about the equality of treatment means?

Consider the following US treasury rate table expressed in percentage.

What is the the yield to maturity of a 2 year 5% bond with annual payments? (Hint: Use the General Formula to find the price first, then compute the yield to maturity)

1. CPK and SGOT tests are used in the diagnosis of myocardial infarction (MI). When the CPK test is given to a patient who does not have a MI, the probability of a negative finding (i.e. its specificity) is 0.6. The probability that the SGOT test will be negative for a non-MI patient is 0.7. When both tests are given to a non-MI patient the probability that at least one is negative is 0.9. For a non-MI patient who has both tests:

1. What is the probability that both tests are negative?
2. What is the probability that both tests are positive?

Hints: (1) Answer is not 0.12 -- tests are not to be assumed to be independent.

           (2) Using 2-by-2 table to structure your calculations can help.

3. What is the probability that at least one test is positive?

An uninsulated steam pipe with a wall temperature of 140 °C passes through a room of quiescent air of 14 °C. The pipe is horizontal and has an outer diameter of 20 cm. Find the following (include units if needed):

a) The temperature to evaluate properties = __________

b) The Rayleigh number, _____________

c) The heat transfer coefficient, h = ____________

d) q’ along the pipe = ________________

e) The pipe diameter is decreased by 50%. If all other parameters remain the same, will the heat transfer coefficient increase, decrease or stay the same? ___________. Justify your answer mathematically. For air: ν = 20.92 x 10-6 m2/s, α = 29.9 x 10-6 m2/s, 30 x 10-3 W/m·K, Pr = 0.7

Batch and continuous biomass production Pseudomonas methylotrophus is used to produce single cell protein from methanol in a 1000-m pressure-cycle airlift fermenter. The biomass yield from substrate is 0.41 gg-, Ks is 0.7 mg 1", and the maximum specific growth rate is 0.44 h. The medium contains 4% (w/v) methanol. A substrate conversion of 98% is desirable. The reactor may be operated in either batch or continuous mode. If operated in batch, an inoculum of 0.01% (w/v) is used and the downtime between batches is 20 h. If continuous operations are used at steady state, a downtime of 25 days is expected per year. Neglecting maintenance requirements, compare the annual biomass production achieved using batch and continuous reactor

(1 point) A manufacturer of electronic kits has found that the mean time required for novices to assemble its new circuit tester is 2.9 hours, with a standard deviation of 0.7 hours. A consultant has developed a new instructional booklet intended to reduce the time an inexperienced kit builder will need to assemble the device and the manufacturer needs to decide whether or not to send out the new booklet.

The testable hypotheses in this situation are

H0:μ=2.9

vs HA:μ<2.9

.

1. Identify the consequences of making a Type I error.
A. The manufacturer sends out a helpful instructional booklet.
B. The manufacturer does not send out a helpful instructional booklet.
C. The manufacturer does not send out an unhelpful instructional booklet.
D. The manufacturer sends out an unhelpful instructional booklet.

2. Identify the consequences of making a Type II error.
A. The manufacturer does not send out a helpful instructional booklet.
B. The manufacturer sends out a helpful instructional booklet.
C. The manufacturer does not send out an unhelpful instructional booklet.
D. The manufacturer sends out an unhelpful instructional booklet.

To monitor the assembly time of inexperienced kit builders using the booklet, the manufacturer is going to take a random sample of 14 novices and calculate the mean time to assemble the circuit tester. If it is less than 2.7, they will send out the new instructional booklet. Assume the population standard deviation is 0.7 hours.

3. What is the probability that the manufacturer will make a Type I error using this decision rule? Round your answer to four decimal places.

4. Using this decision rule, what is the power of the test if the actual mean time to assemble the circuit tester is 2.75 hours? That is, what is the probability they will reject H0 when the actual average time is 2.75 hours? Round your answer to four decimal places.

Assume that the differences are normally distributed. Complete parts ​(a) through ​(d) below.

​(a) Determine

d Subscript i Baseline equals Upper X Subscript i Baseline minus Upper Y Subscript idi=Xi−Yi

for each pair of data.

​(Type integers or​ decimals.)

​(b) Compute

d overbard

and

s Subscript dsd.

d overbardequals=negative 1.650 −1.650

​(Round to three decimal places as​ needed.)

s Subscript dsdequals=1.605 1.605

​(Round to three decimal places as​ needed.)​(c) Test if

mu Subscript dμdless than<0

at the

alphaαequals=0.05

level of significance.

What are the correct null and alternative​ hypotheses?

A.

Upper H 0H0​:

mu Subscript dμdless than<0

Upper H 1H1​:

mu Subscript dμdequals=0

B.

Upper H 0H0​:

mu Subscript dμdgreater than>0

Upper H 1H1​:

mu Subscript dμdless than<0

C.

Upper H 0H0​:

mu Subscript dμdless than<0

Upper H 1H1​:

mu Subscript dμdgreater than>0

D.

Upper H 0H0​:

mu Subscript dμdequals=0

Upper H 1H1​:

mu Subscript dμdless than<0

Fixed acidity - Volatile acidity - Citric acid - Residual sugar -Chlorides

       7.4                         0.7                       0                      1.9                    0.076

      7.8                         0.88                     0                      2.6                    0.098

      7.8                         0.76                   0.04                   2.3                    0.092

      11.2                       0.28                   0.56                   1.9                    0.075

      7.4                         0.7                       0                      1.9                    0.076

      7.4                         0.66                      0                     1.8                    0.075

      7.9                         0.6                     0.06                   1.6                    0.069

      7.3                         0.65                      0                     1.2                    0.065

      7.8                         0.58                   0.02                   2                       0.073

      7.5                         0.5                     0.36                   6.1                    0.071

      6.7                         0.58                   0.08                  1.8                     0.097

      7.5                         0.5                     0.36                   6.1                    0.071

      5.6                         0.615                    0                     1.6                    0.089

      7.8                         0.61                   0.29                   1.6                    0.114

      8.9                         0.62                   0.18                   3.8                    0.176

      8.9                         0.62                   0.19                   3.9                    0.17

      8.5                         0.28                   0.56                   1.8                    0.092

      8.1                         0.56                   0.28                   1.7                    0.368

      7.4                         0.59                   0.08                   4.4                    0.086

      7.9                         0.32                    0.51                  1.8                    0.341

      8.9                         0.22                    0.48                  1.8                    0.077

      7.6                         0.39                    0.31                  2.3                    0.082

      7.9                         0.43                    0.21                  1.6                    0.106

      8.5                          0.49                   0.11                  2.3                    0.084

      6.9                           0.4                     0.14                   2.4                 0.085

      6.3                           0.39                   0.16                   1.4                   0.08

1. For the data on 26 red wines given above, conduct the following analysis:

i. Provide five-number summary i.e. the minimum, 1st quartile, median, 3rd quartile, and maximum value for fixed acidity. Arrange them in increasing order on a straight line, draw a box plot and interpret what it means.

ii. Calculate the correlation coefficient between fixed acidity and volatile acidity and between residual sugar and chlorides. Comment on the strength and direction of association for the two variable pairs.

iii. What can be stated about the cause-effect relationship between fixed acidity and volatile acidity, based on the correlation coefficient score?