bio stats exam 2
Within-group variance
(also called the error) is the mean squared deviation of the observation - the diet mean. (sum of squared deviations from each observation across the whole sample and the treatment means) Under the null hypothesis we would expect the variances to be very close (or equal)
Among-Group variance
(also called treatment variance) is the variance within the groups or treatments
ANOVA testings
) ANOVA is probably the most versatile techniques of statistical inference (hypothesis testing). We use ANOVA (analysis of variance) to analyze the effect of different groups (a categorical variable) on a measurement variable. If your research question is somewhat like: "Are the means of these 3(or more) groups all the same?" you will use ANOVA.
a paired t-test or matched pair MUST have:
-a 1 population with 2 observations -a continous variable -create a new variable difference -the t-distrubution test in the null hypothesis
getting a t crit 1 tailed t-test
-all values in the t PDF table are positive -look at the operator in the alt. hypothesis to determine when graphing
matched pair t-test uses...
-assumptions of the test -continous variable -distribution of variable (differences) is approx. ND -each individual is measured 2x (before and after treatment) and (matched pairs of an individual)
properties of t-distribution
-degrees of freedom (df) (sample size n-1) -sample mean -sample standard deviation -centered at 0 and symmetrical -area under the curve is 1 -asymptotic tails
the t-distribution is used when
-dont have normally distributed data -dont know the population variance -dont have 30+ observations
to compare t-crit to t-stat you...
-draw a t-distribution -mark on the distribution where the t-crit is -mark on the distribution where the t-stat is -reject H0 if the t-calc is within the "rejection zone"
Nonparametric Test: if you don't meet the assumptions of the test...
-your data is ranked -the distribution is not normal
possible t-distribution alpha levels used
0.05 0.01 0.1
T tests
1 population mean t-test -non directional (= in null hypothesis) -directional (<= >= in null hypothesis matched pair t-test independent samples t-test
I want to know if my sample of 20 Guinness bottles have the alcohol volume required by Sir Guinness of 4.8% Alcohol content/ (12oz). To test this I would use the following test:
1 population t-test
Suppose FDA NOW states that vit. X must contain at least 100 units of vit. X per pill. your company can sell vit. X supplements if they have 100 units or more (but not less than 100 units) A sample of 50 pills is collected, sample mean is 100.5 units, sample standard deviation is 2.19 units per pill.
1 tail 1 t test A different Ha H0: u>/ 100 units Ha: u< 100 units (DOESNT HAVE </) only really testing the null H0 sample mean= 100.5 units stand. dev= 2.19 sample= 50 pills standard error of mean Sxbar = .309 find the critical t = df= 49 (sample size 50-1); alpha =0.05. tcrit = 2.011 (see above) Df40/α 0.1 = 1.684 + df60/ α 0.1 = 1.671/2 = -1.677 (- because we are looking in the "less than tail) t-calc= 1.62 draw t-distribution -t-crit to the left (-1.677) -middle = 0.05 -t-cal to the right (1.62) -rejection zone is past 2.01 -1.62 no where near rejection zone --> failed to reject null hypothesis -we can conclude that the average amount of vit. A content is greater than or equal to 100 units -we can sell it with that 5% risk
1 way ANOVA VS Blocking ANOVA
1 treatment or 1 source of variance Adding another source of variance
research and hypotheses
1) formulates -research question -informed by literature 2)testing -research hypothesis -a statement of a relathionship 3)benchmark for measurement -null hypothesis
Assumptions of ANOVA
1. Each of the groups is a random sample from the population of interest 2. The measured variable is continuous (or discrete, with a large range of values) 3. The error variances are equal 4. The variable is approximately normally distributed
steps in hypothesis testing
1. State your null hypothesis 2. State your alternative hypothesis 3. Set your alpha α (we usually use 0.05 other possible values are 0.10, 0.001) 4. choose the correct test! T-test (t-test: 2, tail, 1 tail, ANOVA) 5. find the critical t (crit) for critical F(crit) using your alpha and degrees of freedom (using critical area tables) 6. Compute the test statistic (you will need to know which test statistic, the standard error, etc) 7. Locate the test statistic on the distribution (graphing is suggested) 8. Reject or FAIL to reject the null hypothesis a. If your test statistic is less than the critical value fail to reject the null hypothesis b. If your test statistic is greater than the critical value reject the null hypothesis and support your alt. hypothesis 9. Create a graph to display your results.
inferential stats
1. estimation (confidence intervals) 2. hypothesis testing
Probability Distribution Function process
1. find hypotheses 2. alpha value 3. choose test 4. find critical value (F distri) 5. test statistic (This is a 2 - step process where you calculate the variance among the groups (associated with the diet/treatment) and the variance of the total sample) 6.Compare the F-crit with the F-calc 7. Decision: Reject the null hypothesis or Fail to reject the null hypothesis Interpretation: explain this to my mother in law!
hypothesis testing requires
1. null hypothesis 2. alt. hypothesis 3. null tested with 1PDF (z, t)
Does the ratio of length to width of root hair cells differ in two species of plants (A and B) of the same genus? Plants of each species were measured using random samples from each species. Assume the variable is normally distributed. species A : n= 12 xbar= 1.28 s2= 0.112 species B: n= 18 xbar= 4.43 s2= 2.072
1. 𝐻0: 𝜇𝑎 = 𝜇𝑏 2. 𝐻𝑎: 𝜇𝑎 ≠ 𝜇𝑏 3. This is a 2 tailed t-test for independent samples 4. α = 0.05 5. T-critical value: use the df of the smaller sample (df= 12-1=11) Critical value for 𝑡(0.05,11) = 2.201 6. t-calc: substitute the values from the table above into the t-equation: t=(1.28-4.43)/square root of (0.112/12) + (7.072/18) t=-4.967 7. Compare the t-calc with the t-crit: since this is a 2 tail t-test you can use the absolute value of the tcalc (4.967) tcrit 2.201 on both sides, and t-cal is in rejection zone 8. Decision: Reject the null hypothesis Interpretation: There is a statistically significant difference in the ratio of length to width between the plant species of A and B, at the 0.05 level.
We used a _____ for independent samples when examining the difference between the populations means for 2 groups; we use _____ when examining a difference of means between 2+ groups
2 sample t-test/ANOVA
Suppose you have conducted a 2 tail t-test for a single population mean. You have sampled 30 birds nests across the UNH campus and need to test your null hypothesis at the alpha level of 0.05. What is your t critical value?
2.045
You are a researcher here at UNH examining the balance of fertilzer to use on tomatoe plants grown in a high tunnel (type of greenhouse). You have three different fertilizers to test and you are also looking at 10 different soil types that were previously tested. Therefore, each fertilizer application will be used on individual tomatoe plants in each of the 10 different soil plots. You will use a block design ANOVA for your analysis, with the soils being your blocks. What is the F-crit for Blocks (fertilizer type) in your analysis? Use alpha of 0.05 and the ErrorDF closest to your ErrorDF
2.39
How many null hypotheses will you have when using a Factorial Design ANOVA?
3
You are a researcher here at UNH examining the balance of fertilzer to use on tomato plants grown in a high tunnel (type of greenhouse). You have three different fertilizers to test and you using 10 plots per fertilizer type. What is the F-crit for treatment (fertilizer type) in your analysis? Use alpha of 0.05 and the ErrorDF closest to your ErrorDF
3.35
You have been asked to test a difference of average (corn) yield from 3 different seed varieties. The appropriate test to use for this research question is:
ANOVA explanation: Anytime you are testing a difference of means from more than 2 groups (2+) you should use ANOVA
You are a researcher here at UNH examining the balance of fertilzer to use on tomatoe plants grown in a high tunnel (type of greenhouse). You have three different fertilizers to test and you are also looking at 10 different soil types that were previously tested. Therefore, each fertilizer application will be used on individual tomatoe plants in each of the 10 different soil plots. You will use a block design ANOVA for your analysis, with the soils being your blocks. If you rejected the null hypothesis of no difference between treatment means, but you failed to reject the null hypothesis of no difference between soil types/block means. How would you explain these results?
At least one fertilizer type has a significantly different average yield and there is no statisticaly significant difference in average yield of tomatoes across the soil types.
ANOVA blocking rules
Blocks: are homogenous groups that represent the source of variation we regard as extraneous we wish to remove (or partition out) Treatments: assigned at random to the experimental units within each block; each treatments appears in every block AND each block receives multiple treatments The total number of observations (nt) = number of treatments x the number of blocks Observations sometimes denoted as: Xbt Where b is the block the observation point is from t is the treatment that the observation is given Each block will be given each of the treatments -If there are 10 blocks and 3 treatments; each block will have 3 treatments for a total of 30 observations.
calculate a confidence interval for this hypothesis test species A : n= 12 xbar= 1.28 s2= 0.112 species B: n= 18 xbar= 4.43 s2= 2.072
CI formula for t-test of independent samples with pooled variance (xbar a - xbar b) +- (t0.05,k) *square root of (s2a/na + s2b/nb) K = the degrees of freedom for the smaller of n-1 (smaller sample size) 95%Cl = (4.43 − 1.28) ± 2.201 *square root of (0.112/12) + (7.072/18) 95%Cl =3.15 ± 2.201 ∗ √0.4022 95%Cl =3.15 ± 1.40 = 1.75, 4.55 Interpretation: With 95% confidence the mean difference between group B and A in their root hair cell length-width ratios is at least 1.75 and as great as 4.55.
CI formula -t is:
CI(0.95) =xbar +-(t0.05,n-1 *Sxbar) Sxbar= s/square root of n
Tukey test equation
CV = q (square root of MSe/n) CV = critical value q= critical values of q probability distribution function, found in table A.7 (tukey test) MSe= Mean square for error (taken from the ANOVA table) n= sample size
1 tail t-test
DOES matter which tail t-stat faills H0: u</ u0 H0: u /> u0
what is an example of a independent samples t-test
Do Resting heart rates differ between males and females?
Mann-Whitney U Test
Examines if two samples could be drawn from identical populations. The two population distributions are of the same shape Random Samples drawn from 2 populations Sample sizes need not be equal The two samples are drawn from populations with identical medians
In a block design ANOVA what probability distribution function (pdf) will you use to test the null hypothes(es)?
F distribution
When you suspect an interaction between treatments the appropriate test to use the
Factorial Design ANOVA
F-calcs for Factorial Design ANOVA
Fcalc Treatments: Treatment MS/Error MS Fcalc Blocks: Block MS/Error MS Fcalc Interaction: Interaction MS / Error MS
If you reject the null for a simple one way ANOVA (means are different) you...
Follow with a Tukey to determine a critical value (of difference)
Tukey application
From the Tukey equation get CV Compare absolute value difference between group pairs Treatment 1 - treatment 2 Treatment 1 - treatment 3 Treatment 2 - treatment 3 Groups with difference > CV are significantly different
one tail 1 t-tests: if you are testing to see if something is GREATER than, what do your equations look like?
H0: u /< a value Ha: u > a value (DOESNT HAVE </) signs also switch
Suppose you are a student researcher working on an Ecology research project. You have been asked to sample the nests of the Eastern Wood Pewee bird at various sites surround the UNH campus and woods. Previous research suggests that for this species to have a healthy, sustained population an average of 4 eggs/nest is required. Choose the correct null hypothesis to test whether the population of Eastern Wood Pewee surrounding UNH is considered healthy/sustained.
H0: u=4
null hypothesis equation
H0: u=u0
EXAMPLE: 1 SAMPLE 2 T TEST FDA requirement for Vit. A is 100 units A sample of 50 pills is collected, sample mean is 100.5 units, sample standard deviation is 2.19 units per pill. Do the vit. A pills produced at ____ have average of 100 units?
H0: u=u0 Ha: u doesn't = u0 100.5=100 (null) it is 2 tailed t-test because we were willing to accept a t-stat below 100units or above 100 units. t-crit = 2.01 (taking degree of freedom of 40 and 60 average) t-calc= 100.5 (sample mean) - 100 (hypothesized mean value) /(standard error 2.19/ square root of 50) =0.309 t-calc= 1.62 draw t-distribution (make graph upside down u) on left 2.01, middle line, right 2.01 1.62 note, and you will see that is does not fall in the rejection zone (past 2.01) --> failed to reject null hypo. this means that our sample of vit. A is the equilvanent value of 100 units
one tail 1 t-tests: if you are testing to see if something is LESS than, what do your equations look like?
H0: u>/ a value Ha: u< a value (DOESNT HAVE </)
independent samples t-test hypotheses:
H0: ua = ub Ha: ua doesnt = ub
Hypothesis Statement for ANOVA
H0: ua = ub = uc Ha: ua ≠ ub ≠ uc
Healthy populations of mythical elephant birds have newborn chicks that weigh, on average, 8.0kg. We expect that chicks from populations impacted by disease to weigh less. Suppose we visit a new population and sample 20 randomly selected chicks. Determine which hypothesis to use, write the null and alternative hypotheses, formula used to calculate the test statistic, and find the critical value.
H0: μ ≥ 8.0 vs. Ha: μ < 8.0 t = (7.94 - 8.0)/(0.2037/√20) = -1.3176, t-critical = -1.729, p > 0.1, fail to reject H0, no evidence that the mean weight is less than 8.0 kg
Mann-Whitney U Test equation
H0: 𝜃1 = 𝜃2 𝐻𝑎: 𝜃1 ≠ 𝜃2 MEDIAN (θ, theta) because the median is a good measure of central tendency when the distribution is skewed!
alternative hypothesis
Ha: u doesn't = u0
F calc versus F crit rule
If F calc < F crit, Fail to Reject Ho
Interaction (T*B):
Interaction mean - treatment mean - block mean + overall mean
Factorial Design ANOVA
Main Effects: this refers to the effect of the treatments and blocks. Interactive Effects: refers to the interaction of treatments*blocks
Objective of Blocking
Remove any variation from the error term that could be attributable to the blocks while assuring that treatment means will not be affected by block effects
Power of the test is influenced by:
Significance Level Difference between "true" value of parameter and hypothesized value (effect size) Sample Size (n)
data on the resting and post exercise pulse rates were collected for 8 individuals between ages 19-22 years. we wish to know if there is a difference in pre and post exercise?
Test: matched pair hypotheses: H0: u (post-pre) = 0 Ha: u (post-pre) don't = 0 mean: 35.625 Stand. dev= 23.616 Stand. error: 8.38 alpha: 0.05 tcrit: (0.05, 7) = 2.365 tcal: xbar - 0/ sigma xbar tcal: 35.6-0/8.38 = 4.26 make graph t-crit both sides 2.365 t-cal past 2.3 on right, on regection zone --> reject the null hypothesis, supporting Ha there is a difference in heart rate before and after exercise in individuals aged 19-22 years, at the significance level of 0.05.
When examining the yield of corn from three different seed varieties grown in 10 different soil types; a factorial deisgn ANOVA is used to examine:
The effect of seed types on corn yields and the effect of soil types on corn yield but factorial is particulaly useful in looking at the interaction of 2 factors
Error (residual) deviation
Total deviation - treatment deviation - Block deviation
If you have tested for a difference of means across three groups and you find that you REJECT the null hypothesis supporting your alternative hypothesis that at least one of the groups has a significant difference. To determine which group(s) significantly differ you can use which test:
Tukey test explanation: you can use the Tukey test to estimate a critical value that you can then compare with the absolute difference between group means. i.e. compare the critical value from Tukey equation with xbar1-xbar2= xbar2-xbar3= xbar3-xbar1= If any (or all) of these absolute differences exceed the CV you know that pairing has a significant difference
what happens to the alpha values for a one-tail 1 t-test?
When we are using a 1 tail t-test reading table A.2 requires an additional step of multiplying the alpha value by 2. For example, our alpha value is 0.05, we determine we are using a 1 tail t -test; therefore, we use an alpha of 0.10.
what value of t EXCLUDES 0.05 of the the t-distribution with df=4; alpha 0.05
With DF = 4 , α = 0.05 (0.025 in each tail) Critical value of t = 2.776 ± 2.776 is the value of t that excludes 0.05 of the t - distribution -OR includes 95% of t distribution
Block Deviation
X_(b(bar))-X_barbar (block mean - overall mean)
Treatment Deviations
X_(t(bar))- X_barbar (treatment mean - Overall mean)
What if my degree of freedom is not on the chart?
You will notice that table A.2 does not include all of the possible degrees of freedom. In these instances you can take the average of the critical values that your df falls in. Example: Df= 49 alpha = 0.05 To find the critical area take the average of df=40 and df=60 at alpha 0.05= 2.021 + 2.000/2=2.0105 or 2.011
theory
a hypothesis that has general or widespread appeal to events in the physical universe. a theory cannot be proven "true" but a false theory can be proven "false". the same holds for hypotheses.
Block mean
a mean for each block
Treatment mean
a mean for each treatment
the alternative hypothesis for the treatments in a block design ANOVA with 3 treatments is described as:
at least one treatment differs significantly
In the randomized block design ANOVA why do you have 2 F-calcs?
because there are 2 null hypotheses to test explanation: we only test the null hypothesis. In the block design ANOVA we have 2 null hypotheses to test therefore we have 2 Fcalcs!
ANOVA table: how to find df
between/among= #n-1 within= #n-total df total - n-1
ANOVA table: how to find F calc
between/among= MS n/ MS error
ANOVA table: how to find MS
between/among= SSn/df within= ss Error/error df
Independent Samples t-test
comparing 2 population means Samples are collected randomly from 2 populations of interest The variable "response" measured on a continuous scale; or if discrete must assume a large range of possible variables The variable "response" is approximately normally distributed
matched pair hypothesis (equation)
d= difference between paired values d= x1=x2 x1 is the value of X in first data x2 is the value of X in the second data set
t-distribution chart
degrees of freedom (df) to the left alpha two tailed on the right
what is an example of a matched paired t-test?
determining the effectiveness of medication --> blood pressure
critical values of F distribution chart
df1 (top) = number of groups -1 df2 (left side) = total observations - treatments //total df-group df df2 example: 15 total observations and 3 diets (treatments). df error = (15-3) 12
2 tail t-test
does not matter is t-stat is in either tail H0: u=u0
mean for a single population
done this using Z- normal distribution (known sigma) but what if we dont know the mean or sigma? --> t-distribution
If p-value is > alpha you can
fail to reject the null hypothesis
T or F: The FDA has established that the concentration of a certain pesticide in apples may not exceed 10ppb. A random sample of 50 apples from a major orchard had an average pesticide content of 10.03ppb with a standard deviation of 0.12ppb. In finding the tcritcal value the degree of freedom you will use is 9.
false
T or F: When using the t-distribution you need to know the df. df = n
false
T or F: You have conducted a (2 tail) t -test to test whether the population of Eastern Wood Pewee have an average of 4eggs/nest. You got a t-crit of 2.045 and a t-calc of 1.87. Your decision is to Reject the Null Hypothesis.
false
T or F: You set your alpha value at 0.05 and get a p-value of 0.07 from your hypothesis test; you should reject the null hypothesis.
false
T or F: Pvalues are calculated using the P distribution function (pdf)
false explanation: There is no p-distribution! P values are calculated using the distribution of the test statistic (t, F) and the degrees of freedom
T or F: You have rejected a one-way (simple) ANOVA that tested the difference of means in the blood pressure of python snakes given 3 different types of medication (medication A, B, and control). There were 5 pythons in each medication type for a total of 15 pythons in the experiment. The value of q you will use in your Tukey test is 3.67.
false explanation: q=3.77 use 3 for groups (columns) and df error= total df (14) - treatment df (2) =12
T or F: The hypothesis that we test in stat. analyses is the alternative hypothesis.
false we only test the null hypothesis
T or F: If you are unable to assume that the distribution of a given variable is normally distributed to create a confidence interval for the population mean of that variable you should use the Z distribution.
false explanation: you can't assume that the distribution is normally distributed you should use the t-distribution
Choose the best hypothesis test: You are conducting research on the effects of salinity on larvae survival of Gasterosteus aculeatus. You set up an experiment with 2 different mesocosms (really large fish tanks) with a different level of salinity and in each mesocosm you have 20 Gasterosteus aculeatus larvae in each mesocosm. You will calculate survival as days since hatching.
independent samples t-test
type II error
is failing to reject the null hypothesis when in fact it is false. we call this error beta
Total Variance
is the variance for all observations across all groups.
If you wanted to test if Guinness has an effect on "beer goggles," hypothesized to occur to individuals after the consumption of alcohol that affects the perception of beauty, which would be the most appropriate test to use?
matched pair t-test using a 1 tail test
T-test for independent samples: degrees of freedom
na -1 OR nb-1 you choose whichever is the smaller value of the two
hypothesis can be tested by:
observation/ experiment
what is the best definition or interpretation of a p-value
p<0.05 small probability of getting the results observed under the assumption of the null hypothesis
estimation
refers to the interval within which a parameter (population) falls at some specified probability
statistical inference
refers to the process of making a conclusion about a population based on a sample taken that population -2 areas of inference: hypothesis testing and estimation
If p-value is < alpha you can
reject the null hypothesis
A type I error is:
rejecting the null hypothesis when it is actually true explanation: if your decision from your test statistics leads you to reject the null hypothesis when in reality the null is true you have committed a Type I error
type I error
rejecting the null hypothesis when it is in fact true. this error is called a and we set this value before conducting the statistical test
CI formula -t is used to
sample to estimate Cl for population mean you need Xbar and S (sample standard deviation)
As the sample size increases, the standard error of the mean gets:
smaller explanation: As the sample size gets larger, the SE of the mean gets smaller, and therefore our estimates become more precise.
independent samples t-test equation
t = (xbar a - xbar b) - (ua -ub) / square root of (s2a/na + s2b/nb) xbar= sample mean of population (ua - ub) = zero square root portion: standard error for the difference between means and is based on a pooled estimate of variance
The formula for a matched pair t-test tstat is:
t = barx d / Sbar d D = refers to the variable "difference"
You have been asked to test a difference of average (corn) yield from 2 different seed varieties. The appropriate test to use for this research question is:
t-test for independent samples explanation: 2 populations = independent samples t -test
alpha
the chosen strength of evidence needed to reject the null hypothesis. if the p-value is less than a, then reject the null hypothesis
alternative hypothesis
the opposite of the null; it represents all the other possibilities but is closely related to the research question
p-value
the probability of obtaining a test statistic as extreme (or more extreme) if the null hypothesis was true
the "grand mean" or Overall mean or xbarbar is:
the sum of all observations / n
null hypothesis (H0)
this is the hypothesis that is tested. depending on the research question this hypothesis is often testing "no effect" or the expectation from some specific probability model
single population mean t-test is used:
to estimate population mean for a single population - 1 tail or 2 tail t-test
T or F: A matched pair t-test is a one population t-test.
true
T or F: If the lines in an interaction plot are crossed this would demonstrate an interaction between treatments and blocks.
true
T or F: The F-distribution (critical area table) has 2 degrees of freedoms because it examines the ratio of variance due to treatment(s) and the error variance.
true
T or F: The null hypothesis for an independent samples t-test is: mu subscript A equals mu subscript B
true
T or F: The null hypothesis of an ANOVA tests that there is NO significant difference between groups/treatments. If you reject the null, you can support the alternative hypothesis that at least one group differes signifanctly, but you don't know which group(s) differ significantly.
true
T or F: You have calcuated a one-way (or fixed) ANOVA to test for a difference of average corn yields across 3 different seed varieties. Your Fcalc is 5.259 and your Fcrit is 3.35. You should reject your null hypothesis of no difference in corn yields between seed varieties.
true explanation: You had an F-crit of 3.35, and an F-calc of 5.249; YOu should reject the null hypothesis and conclude that at least one seed variety differs.
T or F: You have calcuated a one-way (or fixed) ANOVA to test for a difference of average corn yields across 3 different seed varieties. Your Fcalc is 5.259 and your Fcrit is 3.35. You should reject your null hypothesis of no difference in corn yields between seed varieties.
true explanation: You had an F-crit of 3.35, and an F-calc of 5.249; You should reject the null hypothesis and conclude that at least one seed variety differs.
T or F: If you were to test whether or not the Vitamin A, manufactured at the company you are employed by, has 100 units you would use a 2 tailed t-test.
true explanation: You want to know if the vitamin A has 100 units - you are not specifying if Vitamin A has less than or greater than 100 units
T or F: You have completed an independent samples t-test on the differences of average yield (spinach) between 2 seed varieties. Your decision is to reject the null hypothesis at the 0.05 level. You then calculate a 95% Confidence interval and find the difference in spinach yield to be at least 12 ounces and as much as 35 ounces. Your confirm that your CI supports your hypothesis test.
true explanation: the null hypothesis for a difference of means from 2 population states that ('mu'1 - 'mu'2 = 0), you rejected the null hypothesis that supports the alt. hypothesis that ('mu'1 - 'mu'2 not equal 0). Your CI interval does not include a zero so this does support your hypothesis test.
t-calc (t-stat) equation
ts = xbar - u0/ s/square root n ts= t statistic xbar= sample meet u0= our hypothesized value of population mean s/square root n = sample standard error of the mean
If you have tested for a difference of means across three groups and you find that you REJECT the null hypothesis supporting your alternative hypothesis that at least one of the groups has a significant difference. To determine which group(s) significantly differ you can use which test:
tukey test explanation: you can use the Tukey test to estimate a critical value that you can then compare with the absolute difference between group means.
In a randomized block design ANOVA, how many null hypothesis will you have?In a randomized block design ANOVA, how many null hypothesis will you have?
two 2-1 null for the treatments; 1 null for the blocks
hypothesis testing
using data to evaluate the validity of a particular idea (or hypothesis)
statistical significance
we often say a statistic is statistically sign. at the associated p-value if this p-value is less than the set alpha level. in other words, if you can reject your null hypothesis you can report that your test statistic is statistically sign. the ____ p-value
a random sample of 20 mosquito fish was collected, the length of fish was observed (mm). the average length is 21mm, with a standard deviation of 1.76mm
xbar = 21mm stand dev. (s) = 1.76 sample size (n) = 20 S xbar = 1.76/square root of 20 = 0.394 CI(0.95) =xbar +-(t0.05,n-1 *Sxbar) CI(0.95) = 21+- (t.05,19 *0.394) value of t you have to look at chart (df = 19 and alpha = .05 --> 2.093) answer: 20.175mm<u<21.8mm we would expect with 95% confidence that the true length of the population falls between 20.17mm and 21.8mm
