Question

You will then computea statistic from the raw date. You can find the p value associated with this statistic in the same manner by locating where would le within the row that corresponds to the appropriate degrees of freedom. If this value is below the a value, you will reject the nutrypothesis in favor of an alternative hypothesis, which states that the means are significantly different DATA ANALYSIS AND HYPOTHESIS TESTING Art wure of a well-designed experiment la replication, which reduces the possibility that the results obtained occurred by chance alone in this in each potism Independent experimental Hlustrating the growing conditions for a given treatment Every effort was made to keep environmental conditions such as exposure to sunlight and moisture identical among pols, but in reality this is impossible. By using several replicates of each treatment, we increase our confidence that the results reflect these unique treatments, rather than chance variation. The data given to you in your worksheet represents the average mass of individual plants within a pot. The average mass in the data table representa replicate pots for each treatment average mess of individual plants 22.327 4.604 4032 7.172 5.399 5202 critical values of the distribution P values 0.50 0.20 0.10 905 0.005 0002 0.001 1 1.000 3.078 6.314 12.706 31621 657857 27 321 318,309 661 2 0.816 18 2920 4.303 6.965 3 9.925 14.000 31.599 0.765 1.6.38 2.353 3.182 7.453 10.215 12924 4 0.741 1.539 2.132 2.778 3.747 5.500 8610 0.727 1.476 2015 2571 3.365 4.773 6.169 0.718 1.440 1.943 2447 3.143 3 707 4317 0711 14:15 185 2.36 2.998 3.490 4029 4.785 5400 2.800 3.366 3.833 4.501 5041 9 1.383 1.833 212 2021 3.250 3690 4 297 4781 10 0700 1.372 1.812 2.228 2.784 3.160 3.501 4587 11 0.697 133 1.796 2201 2.718 3.106 3.497 4025 4437 12 0.695 1.356 1.782 2.170 2681 3.930 4318 13 0.694 1.350 1.771 2.100 2.6503012 3372 14 0.682 1.345 1.781 2.145 3.852 4221 2.624 29773320 4140 15 0.601 1341 1753 2.131 2.6022047 3.20 37334073 16 0.600 1.337 1.746 2.120 2.583 20213252 3.61 4.015 7 0.706 1.100 070 3088 3787 average mass amy replicate pots We will then choose two mean values to compare and use a familiaritical test Students les to determine whether they we wrificantly different. The ruil permessina wo sampelos is that the mom of the two groups wel The afference between group means is gedeflative to both the size and the variability of the samples The formula for the test is a wo. The numerator is the whole value of the difference between the two sample means. The denominator is a more of prevarily of the Marsandard error. This com essentially related signal to noise the difference between the means in the son of an experimental en introduced the variability is noise that may make it hard to see this difference For example you the data from samples for each of treatment groups, you egment has degrees of freedom.6.2Within a value of 0.05, your critical value 82306 As long as your called in a greater than 2300. e corresponding value wallow 0.06, and you can reach nu hypothesis + -statistie SD SD mean 50 standard deviation number of samples EXAMPLE DATA FROM GROUP • Your local plant species radiah . Al above ground plant mass torradhes any was collected from each pet (wheat not weighed) • Mass values to 001 gram were recorded in the table below The best statistici gets large as the means get further apart and gets smaller as the variances decrease of sample se increase. The probability of getting the observed! value assuming the null hypothesis is true is determined using the distribution. The shape of hal darbuon depends on the number of degrees of freedom, which is equal to the sun of the sample size of each group being compared minus 2 (-2) Next, you must choose a critical p value or a value. An a value of 0.05 means that in experiments out of 100 you would find a stistically significant difference between Samole mont by chance alone. You can then find the associated critical value at the intersection of the column corresponding to the a value and the row corresponding to the degrees of freedom. 49 20 Radiah 20 Radish. 20 When 100 100 2 mesos 1720 3980 2860 moddish only 3620 2760 4 mats per dich and 13/2016 1/40.0.97 01/20 - 1 ML

Answer 1

Page No.. 20 rallish: 40 nadlid 7 t- Test Calculations 71082-1.007 0.82 om (0.07) trat = 0.82 = 10.0 0.001370.0054 2. 0819 degrees of freedom -/= 4-1 = 8 x-value 11 0.05 Critical t value = 2.306 p-value Tes Since tal > teritical mull-leypothesis is rejected This means the are Brembently different mass/ plant is lesser in 40 raddish 20 reddish. This is due to plants for limited for resources. case Thus nutrien competition of the

-teal 14.82-1.211 (o.nl 20 raddithi zondla 20 wheat 0.61 16.0013 & 2.0032 03 0.0794 7.68 degrees of freedan - 8 8 0.05 t critical = 2.306 Р >0.00! suppothesis is teal > tcritical null rejected mass /plant in 20 raddish/20 wheat Ergnificantly lower than 20 raddish pot This that wheat also competes with raddish nutrients and affects mass/ plant.

t-cal = 0.21 0.2/ 2.26 0.005 0.8032 0.093 40 was 30 ml dogress of freedom 0.05 Critical to value = 2.306 "0.05 means -X مگر ان وجوده molt bypettiesis is accepted not affected by species competing nutrients. Mars/plant decreases more plants / pot isrespective of raddish (wheat. balrá specifieke competition is shkis.lt to Denter specific competition for effects note 07