Researchers aim to find a topical solution for skin cancer in its early stages to avoid surgery.They believe Curcumin is effective, however, it must be solubilised in a solution as a carrierto be applied on skin. Solvability is one important factor because a solution with highersolvability contains a higher amount of curcumin. For this reason, an experiment wasdesigned by researchers to examine and compare the solvability of 5 popular solutions. Priorto starting the experiment, a company asked the researchers to use and examine their newadditive solution "X" by adding the additive to the solutions. The company claims that theadditive significantly increases the solvability of the solutions. The researchers test thesolvability of their 5 solutions first and then redo all the steps again by adding "X" to theirsolutions. There are 5 replicates for each solution in each test. The results are displayed in thetable below:Withoutadditive "X"Withadditive "X"Water1.11.11.31.41.29.38.110.88.810.75Ethanol108.59.210.2912.29.39.810.511.7Solvability mg/mlCapryol9.459.210.28.99.59.58.79.18.28.2Labrafac21.8524.423.522.6524.12725.828.828.427.1Labrasol52.149.955251.852.549.353.251.148.550.2a) Run a two independent samples t-test in the statistical software package of your choice forbefore and after adding solution "X" and interpret your finding.Do you think this test can tell us anything about using the additive or not?

Note that I used R-software for my calculations. Null hypothesis: Solvability remains same before…

Water 1.1 Without 1.1 additive "X" 1.3 1.4 1.2 9.3 8.1 10.8 8.8 10.75 With additive "X" Ethanol 10 8.5 9.2 10.2 9 12.2 9.3 9.8 10.5 11.7 Solvability mg/ml Capryol 9.45 9.2 10.2 8.9 9.5 9.5 8.7 9.1 8.2 8.2 Labrafac 21.85 24.4 23.5 22.65 24.1 27 25.8 28.8 28.4 27.1 Labrasol 52.1 49.95 52 51.8 52.5 49.3 53.2 51.1 48.5 50.2 ) Run a two independent samples t-test in the statistical software package of your choice for before and after adding solution "X" and interpret your finding. Do you think this test can tell us anything about using the additive or not?

Water 1.1 Without 1.1 additive "X" 1.3 1.4 1.2 9.3 8.1 10.8 8.8 10.75 With additive "X" Ethanol 10 8.5 9.2 10.2 9 12.2 9.3 9.8 10.5 11.7 Solvability mg/ml Capryol 9.45 9.2 10.2 8.9 9.5 9.5 8.7 9.1 8.2 8.2 Labrafac 21.85 24.4 23.5 22.65 24.1 27 25.8 28.8 28.4 27.1 Labrasol 52.1 49.95 52 51.8 52.5 49.3 53.2 51.1 48.5 50.2 ) Run a two independent samples t-test in the statistical software package of your choice for before and after adding solution "X" and interpret your finding. Do you think this test can tell us anything about using the additive or not?

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Week Job Sat Score 1 It is 6% higher It is 12% higher It is 43% higher They are the same 2 3 4 5 6 7 8 3.40 3.60 3.70 3.80 3.90 4.20 4.00 4.10 Absence Rate 0.05 0.04 0.06 0.03 0.04 0.03 0.04 0.05 Quality Score 0.88 0.89 0.90 0.92 0.83 0.91 0.92 0.83 How does the job satisfaction score for weeks 5-8 compare to weeks 1-4? It is 43% lower Output Score 8 10 9 10 8 10 9 Weekly Goals: Job Satisfaction Score ≥ 4.50, Absence Rate ≤ 0.03, Quality Score ≥ 0.95, Output Score ≥ 9 9
SAMPLE NAME pH EC DATE SAMPLED Effluent 1 7.1 80 27/02/2020 Effluent 2 5.1 410 27/02/2020 Effluent 3 5.0 1400 27/02/2020 Effluent 1 7.0 80 25/01/2020 Effluent 2 6.5 730 25/01/2020 Effluent 3 7.1 800 25/01/2020 Effluent 1 7.2 400 24/12/2019 Effluent 2 6.9 700 24/12/2019 Effluent 3 7.2 790 24/12/2019 Effluent 1 7.1 250 23/11/2019 Effluent 2 6.8 750 23/11/2019 Effluent 3 7.0 800 23/11/2019 Effluent 1 7.0 350 25/10/2019 Effluent 2 6.9 810 25/10/2019 Effluent 3 7.1 810 25/10/2019 Effluent 1 7.0 750 28/09/2019 Effluent 2 7.0 350 28/09/2019 Effluent 3 7.0 770 28/09/2019 Effluent 1 7.0 750 25/08/2019 Effluent 2 7.0 350 25/08/2019 Effluent 3 7.0 750 25/08/2019 Effluent 1 7.0 350 27/07/2019 Effluent 2 7.1 680 27/07/2019 Effluent 3 6.8 760 27/07/2019 Effluent 1 7.2 400 27/06/2019 Effluent 2 6.8 650 27/06/2019 Effluent 3 6.7 760 27/06/2019 Effluent 1 7.0 350 26/05/2019 Effluent 2 6.9 750 26/05/2019 Effluent 3 7.0 770 26/05/2019 Effluent…
Anova exercise . Show the step and output using Excel.
Identify the following values. Consider the function shown with a graphical depiction of limƒ ( x = L , including the roles XC of & and d. 201 1.8- 1.6 1.4- 1.2- Y 1.0- 0.8- 0.6- 0.4 0.2- 0.0+T 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 1. 1.6 2. 1.0 E = 3. 0.4 = %3D 4. 0.6 >
urgent pleqase
s1
Full data set O Uncarpeted 13.2 A researcher wanted to determine if carpeted rooms contain more bacteria than uncarpeted rooms. The table shows the results for the number of bacteria per cubic foot for both types of rooms. Carpeted 10.7 14.3 11 6.7 12.8 10 8.4 12.2 9.2 11.1 6.7 11.3 14.1 13.6 8.9 Determine whether carpeted rooms have more bacteria than uncarpeted rooms at the g = 0.05 level of significance. Normal probability plots indicate that the data are approximately normal and boxplots indicate that there are no outliers. State the null and alternative hypotheses. Let population 1 be carpeted rooms and population 2 be uncarpeted rooms. A. Ho: H1 = H2 H1: 41> H2 O B. Ho: H1 = H2 O C. Ho: H1 = H2 H1: H1 H2 Determine the P-value for this hypothesis test. P-value = (Round to three decimal places as needed.)
College Algebra Exponential and Logarithmic Equations Activity The risk of having a car accident increases exponentially as the concentration of alcohol in the blood increases. The risk is modeled by R = 6e12.77x where x is the blood alcohol concentration (BAC) and R, written as a percent, is the risk of having a car accident. According the California Department of Motor Vehicles (DMV) it is illegal for any person to operate a vehicle with a: • BAC of 0.08% or higher, if the person is 21 years old or older. • BAC of 0.01% or higher, if the person is under 21 years old. What Blood Alcohol Concentration corresponds to a 17% risk of a car accident? What Blood Alcohol Concentration corresponds to a 7% risk of a car accident? Use the Blood Alcohol Concentration chart (on back) to determine the likelihood a 140 pound (lbs) female will be in an accident if she has had 5 drinks.
A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 114, and the sample standard deviation, s, is found to be 10. (a) Construct a 96% confidence interval about µ if the sample size, n, is 21. (b) Construct a 96% confidence interval about µ if the sample size, n, is 12. μ (c) Construct an 80% confidence interval about µ if the sample size, n, is 21. (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Click the icon to view the table of areas under the t-distribution. (a) Construct a 96% confidence interval about µ if the sample size, n, is 21. Lower bound: Upper bound: (Use ascending order. Round to one decimal place as needed.)
D2L Hawkes SSO ( X Lesson 4.8 Pra X D2L B.6 Process Pa X y google Yaho X X A Answers - The X y! percent to de X learn.hawkeslearning.com/Portal/Lesson/lesson_practice#! - Save & Exit Practice Lesson: 4.8 Simple and Compound Interest MARTIKA GRC Question 1 of 9, Step 1 of 1 0/9 Correct If the value of your house is $169,000 today and inflation is constant at 10.7% annually, what will be its value, to the nearest thousand dollars, in 7 years? Answer How to enter your answer (opens in new window) Keyboard $ I G answers.com Σ Percent to De
Give Final answer correct which i fill please do fast
-In statistics, the percentage is the same or equivalent to the proportion (p). Percent means per 100. For example, if there are 1,000 residents in a town, and 60% are Democrats, then, on average, 60 out of each 100 are Democrats. You determine the total number of Democrats by converting the percentage to a proportion, .60, and multiplying .60 x 1,000, which equals 600. -To determine a percentage, we divide the part by the whole and multiply by 100. So, if 8 out of 234 seniors are reported to have tried Meth, we calculate the percentage: 8/234 0.0342 and multiply 0.0342 x 100= 3.42%. -Remember: when you multiply by 10, move the decimal point one place to the right3; when you multiply by 100, move the decimal point two places to the right, etc. Answer the questions that follow the Table below Adolescents at three drug treatment centers were surveyed. The questionnaire used consisted of 102 questions and was divided into four sections. The third section asked about past drug use. The…