In an experiment involving the breaking strength of a certain type of thread used in personal flotation devices, one batch of thread was subjected to a heat treatment for 60 seconds and another batch was treated for 120 seconds. The breaking strengths (in N) of ten threads in each batch were measured. The results were 60 seconds: 43 52 52 58 49 52 41 52 56 58 120 seconds: 59 55 59 66 62 55 57 66 66 51 Let μX represent the population mean for threads treated for 120 seconds and let μY represent the population mean for threads treated for 60 seconds. Find a 99% confidence interval for the difference μX−μY . Round down the degrees of freedom to the nearest integer and round the answers to three decimal places.
The 99% confidence interval is
In: Statistics and Probability
In order to determine the possible effect of chemical treatment on women’s hair, an experiment was conducted. The results are given in the accompanying table
| TREATMENT | GROWTH | NO GROWTH |
|---|---|---|
| CHEMICALLY TREATED | 118 | 22 |
| UNTREATED | 120 | 40 |
a) State the null hypothesis and the alternative hypothesis
b) Calculate (and show) the expected frequency for each cell. Place each expected value in parentheses ( ) in the same cell (and to the right) as the observed value to which it corresponds
c) Using the formula: X^2= ∑(O-E)^2/(E) calculate the chi-square value.
d) At the 5% level of significance, what is the chi-square critical value?
e) What decision would you make?
f) Draw a conclusion (in the context of the problem)?
In: Statistics and Probability
In calculating the percent yield for this experiment, which of following the statements is/are correct?
|
1. Just like any percent yield calculation you need to compare grams starting material to grams product. 2. In this unique case we can compare grams to grams since our molecular formula is the same and our mole ratio is 1:1. 3. Just like any percent yield calculation we can use the balanced equation to determine the mole ratio first. Then using this mole ratio we can determine the expected amount of moles of product that could form based on the amount of starting material. We then take the actual amount of moles isolated and divide it by the expected amount of moles calculated to determine a percentage. 4. The mole ratio between Maleic and Fumaric Acid is 1:1. 5. We cannot determine a percent yield, as no reaction has occurred. The reaction is maleic acid reacted in the prescense of HCL to form fumaric acid. More than one can be correct. |
In: Chemistry
An experiment studied the efficacy of using 95% ethanol or 20% bleach as a disinfectant in removing bacterial and fungal contamination when culturing plant tissues. The experiment was repeated 15 times with each disinfectant, using eggplant as the plant tissue being cultured. Five cuttings per plant were placed on a petri dish for each disinfectant and stored at 25°C for four weeks. The observation reported was the number of uncontaminated eggplant cuttings after the four-week storage.
| Disinfectant | 95% Ethanol | 20% Bleach |
| Mean | 3.76 | 4.84 |
| Variance | 2.78099 | 0.17149 |
| n | 15 | 15 |
| Pooled variance 1.47624 | ||
(a) Are you willing to assume that the underlying variances are equal?
Yes
No
(b) Using the information from part (a), are you willing to
conclude that there is a significant difference in the mean numbers
of uncontaminated eggplants for the two disinfectants tested? (Use
μ1 for 95% ethanol disinfectant and
μ2 for 20% bleach disinfectant. Use α
= 0.05.)
State the null and alternative hypotheses.
H0: (μ1 − μ2) < 0 versus Ha: (μ1 − μ2) > 0
H0: (μ1 − μ2) = 0 versus Ha: (μ1 − μ2) ≠ 0
H0: (μ1 − μ2) = 0 versus Ha: (μ1 − μ2) > 0
H0: (μ1 − μ2) = 0 versus Ha: (μ1 − μ2) < 0
H0: (μ1 − μ2) ≠ 0 versus Ha: (μ1 − μ2) = 0
State the test statistic. (Round your answer to three decimal
places.)
t =
State the p-value.
p-value < 0.010
0.010 < p-value < 0.020
0.020 < p-value < 0.050
0.050 < p-value < 0.100
0.100 < p-value < 0.200
p-value > 0.200
State the conclusion.
H0 is rejected. There is insufficient evidence to conclude that there is a significant difference in the mean numbers of uncontaminated eggplants for the two disinfectants used.
H0 is not rejected. There is sufficient evidence to conclude that there is a significant difference in the mean numbers of uncontaminated eggplants for the two disinfectants used.
H0 is not rejected. There is insufficient evidence to conclude that there is a significant difference in the mean numbers of uncontaminated eggplants for the two disinfectants used.
H0 is rejected. There is sufficient evidence to conclude that there is a significant difference in the mean numbers of uncontaminated eggplants for the two disinfectants used.
In: Statistics and Probability
This isn't a homework assignment more of a search for advice. The lab experiment that my group is conducting is to separate two mixtures, a solid-solid and a liquid-liquid mixture. Each mixture has 2 separate components. The solid-solid mixture was easy enough, a quick solvent extraction completely separated the two and soon we will do IR, NMR and melting point determination to try to identify the solids.
The liquids however are another problem all in themselves. The mixture is tinted yellow and is translucent. The two liquids are entirely miscible and there seems to be no obvious phycial way to separate them. There is nothing known about the liquids currently except what we've measured;
The mixture is neutral pH
The mixture has at least one aromatic comound
We've been unable to observe a major polarity difference in the mixture, we had time to run 2 TLC plates, one with polar eluent and one with a mixed 75/25 polar/non-polar eluent. The Rf values have overlapped on both so we believe the compounds have similar polarities, we plan on running a few more with different eluents. This is also why we've been unable to determine whether both compounds are aromatic or only one is.
There is a good chance that the compounds are constitutional isomers of each other based on the hints that our TA has been dropping. We're extremely reluctant to attempt a distillation because as far as we can tell the polarities are extremely similar, as it is directly related to boiling point it is unlikely that the boiling points will be different enough to separate the compounds without a professional distillation set-up. We only have microscale organic kits to work with and previous distillations using the kit resulted in major loss of compound if there is any separation at all.
The rules state we cannot use more than 5mL of the unknown mixture total, for all of the tests we utilize. As a result we cannot afford to lose any to faulty processes like distillation.
The Characterizational techniques we are experienced in using are IR, Melting point, boiling point, refractive index, polarimetry. We are NOT allowed to use Mass Spec. and we can ONLY use NMR after we can prove that we are reasonably confident in the identity of our unknowns.
We're out of ideas so we're hoping for some solid advice and discussion. Thank you!
Tools we have access to are;
TLC
Staining Agents
UV lights
Sep. Funnel
Microscale Distillation (not recommended)
Essentially unlimited glassware
Hot plates and ice-baths
Reagents: HCl, NaOH, Sodium Anhydride, various solvents: toluene, hexanes, ethyl acetate, diethyl ether, water, ethanol, methanol.
I highlighted any relevant information regarding the liquids.
In: Chemistry
A factorial experiment was designed to test for any significant differences in the time needed to perform English to foreign language translations with two computerized language translators. Because the type of language translated was also considered a significant factor, translations were made with both systems for three different languages: Spanish, French, and German. Use the following data for translation time in hours.
| Language | |||
| Spanish | French | German | |
| System 1 | 6 | 13 | 11 |
| 10 | 17 | 15 | |
| System 2 | 4 | 16 | 15 |
| 8 | 18 | 21 | |
Test for any significant differences due to language translator system (Factor A), type of language (Factor B), and interaction. Use .
Complete the following ANOVA table (to 2 decimals, if necessary). Round your p-value to 4 decimal places.
| Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | -value | |
| Factor A | |||||
| Factor B | |||||
| Interaction | |||||
| Error | |||||
| Total |
The p-value for Factor A is - Select your answer -less than .005between .005 and .0125between .0125 and .025between .025 and .05greater than .05Item 21
What is your conclusion with respect to Factor A?
- Select your answer -Factor A is not significantFactor A is significantItem 22
The p-value for Factor B is - Select your answer -less than .005between .005 and .0125between .0125 and .025between .025 and .05greater than .05Item 23
What is your conclusion with respect to Factor B?
- Select your answer -Factor B is not significantFactor B is significantItem 24
The p-value for the interaction of factors A and B is - Select your answer -less than .005between .005 and .0125between .0125 and .025between .025 and .05greater than .05Item 25
What is your conclusion with respect to the interaction of Factors A and B?
- Select your answer -The interaction of factors A and B is not significantThe interaction of factors A and B is significant
In: Statistics and Probability
In an experiment to determine the effects of conventional and reduced tillage agriculture on crop yield for oats, 3 varieties of oats and two levels of fertilization (0.5 and 1 kg/ acre) were examined using conventional and reduce tillage techniques. Twenty 20 x 60m plots were each partitioned into 6 -10 x 20m subplots and subplots assigned at random to receive a combination of oat variety and fertilizer treatment. Ten of these plots were subjected to conventional tillage practices and 10 were subjected to reduced tillage practices. Each subplot was harvested at seasons' end and crop yield is expressed in bushels/acre. Examine the data provided in the data file oats.csv. Report hypothesis tests for all possible effects.
Data provided in the data file oats.csv. :
| tr | fert1v1 | fert1v2 | fert1v3 | fert2v1 | fert2v2 | fert2v3 | |
| 1 | 1 | 55.59555 | 62.28367 | 54.02856 | 52.3082 | 66.11916 | 47.28625 |
| 2 | 1 | 54.83042 | 66.4528 | 58.34965 | 51.20269 | 70.35817 | 59.59235 |
| 3 | 1 | 47.17417 | 53.11721 | 48.0996 | 42.20308 | 56.72854 | 49.01972 |
| 4 | 1 | 49.92158 | 59.02375 | 51.37804 | 47.1327 | 63.70916 | 51.65932 |
| 5 | 1 | 63.2877 | 69.37552 | 67.33111 | 60.75426 | 76.44359 | 64.48788 |
| 6 | 1 | 43.26808 | 51.84271 | 43.80456 | 43.46588 | 58.33783 | 41.24954 |
| 7 | 1 | 53.43448 | 61.23901 | 56.50824 | 55.53189 | 66.73144 | 54.54501 |
| 8 | 1 | 52.09698 | 59.69132 | 56.43002 | 46.99428 | 67.06747 | 57.05954 |
| 9 | 1 | 53.23395 | 59.83314 | 57.81001 | 54.74937 | 64.69777 | 57.40591 |
| 10 | 1 | 43.52501 | 48.36869 | 43.50235 | 40.54583 | 53.5147 | 45.39921 |
| 11 | 2 | 49.76705 | 52.92597 | 51.92696 | 48.5611 | 58.35346 | 48.73186 |
| 12 | 2 | 57.11089 | 60.52846 | 55.08849 | 52.75431 | 63.08562 | 58.9274 |
| 13 | 2 | 57.08448 | 69.46167 | 59.38628 | 59.41141 | 77.39677 | 59.74098 |
| 14 | 2 | 55.65932 | 64.41633 | 55.02598 | 55.76469 | 72.4667 | 56.30465 |
| 15 | 2 | 67.73778 | 72.19191 | 67.87371 | 67.11457 | 77.51533 | 64.34463 |
| 16 | 2 | 63.81949 | 66.31592 | 59.74957 | 63.24728 | 70.01633 | 61.11576 |
| 17 | 2 | 59.57243 | 63.58954 | 59.51328 | 60.33233 | 68.53625 | 59.57449 |
| 18 | 2 | 53.61506 | 60.7735 | 56.80479 | 47.1538 | 64.1669 | 59.46785 |
| 19 | 2 | 61.84216 | 67.61936 | 64.58683 | 57.77677 | 75.37341 | 63.0163 |
| 20 | 2 | 60.88292 | 67.25251 | 60.19583 | 58.40519 | 72.17987 | 57.51454 |
In: Statistics and Probability
Fertilizer: In an agricultural experiment, the effects of two fertilizers on the production of oranges were measured. Thirteen randomly selected plots of land were treated with fertilizer A, and
10
randomly selected plots were treated with fertilizer B. The number of pounds of harvested fruit was measured from each plot. Following are the results.
| Fertilizer A | ||||||
| 523 | 464 | 483 | 460 | 491 | 403 | 484 |
| 448 | 457 | 437 | 516 | 417 | 420 | |
| Fertilizer B | ||||
|
362 |
414 |
408 |
398 |
382 |
|
368 |
393 |
437 |
387 |
373 |
|
Part 1 of 3
Your Answer is correct
Explain why it is necessary to check whether the populations are
approximately normal before constructing a confidence
interval.
Since the sample size is ▼small, it is necessary to check that the populations are approximately normal.
Part 2 of 3
Your Answer is correct
Following are boxplots of these data. Is it reasonable to assume
that the populations are approximately normal?
400 420 440 460 480 500 520 540
360 370 380 390 400 410 420 430 440 450
It ▼is reasonable to assume that the populations are approximately normal.
Part 3 of 3
Construct a 95% confidence interval for the difference
between the mean yields for the two types of fertilizer. Let
μ1 denote the mean yield for fertilizer A. Use the TI-84 Plus
calculator. Round the answers to one decimal place.
| The 95% confidence interval for the
difference between the mean yields for the two types of fertilizer
is
< μ1 - μ2< . |
In: Statistics and Probability
Fertilizer: In an agricultural experiment, the effects of two fertilizers on the production of oranges were measured. Thirteen randomly selected plots of land were treated with fertilizer A, and
10
randomly selected plots were treated with fertilizer B. The number of pounds of harvested fruit was measured from each plot. Following are the results.
| Fertilizer A | ||||||
| 523 | 464 | 483 | 460 | 491 | 403 | 484 |
| 448 | 457 | 437 | 516 | 417 | 420 | |
| Fertilizer B | ||||
|
362 |
414 |
408 |
398 |
382 |
|
368 |
393 |
437 |
387 |
373 |
|
Part 1 of 3
Your Answer is correct
Explain why it is necessary to check whether the populations are
approximately normal before constructing a confidence
interval.
Since the sample size is ▼small, it is necessary to check that the populations are approximately normal.
Part 2 of 3
Your Answer is correct
Following are boxplots of these data. Is it reasonable to assume
that the populations are approximately normal?
400 420 440 460 480 500 520 540
360 370 380 390 400 410 420 430 440 450
It ▼is reasonable to assume that the populations are approximately normal.
Part 3 of 3
Construct a 95% confidence interval for the difference
between the mean yields for the two types of fertilizer. Let
μ1 denote the mean yield for fertilizer A. Use the TI-84 Plus
calculator. Round the answers to one decimal place.
| The 95% confidence interval for the
difference between the mean yields for the two types of fertilizer
is
< μ1 - μ2< . |
In: Statistics and Probability
An experiment consists of randomly rearranging the 10 letters of
the word QUARANTINE
into a sequence of 10 letters, where all possible orders of these
10 letters are equally likely. Find the probability of each of the
following events:
(1) the first three letters include no A’s;
(2) the first three letters or the last three letters (or both) include no A’s;
(3) the fourth letter is the first A;
(4) the first letter and the last letter are the same;
(5) the word ‘QUARANTINE’ is obtained;
(6) the sequence contains the word ‘RAN’.
In: Statistics and Probability