To explain do the data show that dead cicadas increase seed mass.
Answer to Problem 27.43E
The data do not show that dead cicadas increase seed mass.
Explanation of Solution
In the question, it is given the data for the seed masses produced by cicada-fertilized plants and undistributed control plants. Thus, to find if the data show that dead cicadas increase seed mass or not, then we will use the Wilcoxon rank sum test by using software. The hypotheses will be defined as:
Null hypothesis: There is no difference between them.
Alternative hypothesis: The dead cicadas increase seed masses than the control group.
Thus, the result for the test by using the software is as:
Cicada plants | Control group |
0.237 | 0.212 |
0.109 | 0.261 |
0.261 | 0.203 |
0.276 | 0.215 |
0.239 | 0.178 |
0.238 | 0.29 |
0.218 | 0.268 |
0.351 | 0.246 |
0.317 | 0.241 |
0.192 | 0.188 |
0.277 | 0.265 |
0.209 | 0.241 |
0.227 | 0.285 |
0.234 | 0.244 |
0.266 | 0.253 |
0.21 | 0.19 |
0.263 | 0.145 |
0.245 | 0.263 |
0.31 | 0.135 |
0.201 | 0.257 |
0.241 | 0.198 |
0.238 | 0.19 |
0.171 | 0.249 |
0.255 | 0.196 |
0.296 | 0.247 |
0.295 | 0.253 |
0.305 | 0.17 |
0.226 | 0.155 |
0.223 | 0.266 |
0.211 | 0.212 |
0.142 | 0.253 |
0.277 | 0.22 |
0.235 | 0.14 |
0.296 | |
0.217 | |
0.193 | |
0.257 | |
0.276 | |
0.229 |
Label | Data | Rank |
Cicada plants | 0.237 | 34 |
Cicada plants | 0.109 | 1 |
Cicada plants | 0.261 | 52.5 |
Cicada plants | 0.276 | 60.5 |
Cicada plants | 0.239 | 37 |
Cicada plants | 0.238 | 35.5 |
Cicada plants | 0.218 | 26 |
Cicada plants | 0.351 | 72 |
Cicada plants | 0.317 | 71 |
Cicada plants | 0.192 | 13 |
Cicada plants | 0.277 | 62.5 |
Cicada plants | 0.209 | 19 |
Cicada plants | 0.227 | 30 |
Cicada plants | 0.234 | 32 |
Cicada plants | 0.266 | 57.5 |
Cicada plants | 0.21 | 20 |
Cicada plants | 0.263 | 54.5 |
Cicada plants | 0.245 | 42 |
Cicada plants | 0.31 | 70 |
Cicada plants | 0.201 | 17 |
Cicada plants | 0.241 | 39 |
Cicada plants | 0.238 | 35.5 |
Cicada plants | 0.171 | 8 |
Cicada plants | 0.255 | 49 |
Cicada plants | 0.296 | 67.5 |
Cicada plants | 0.295 | 66 |
Cicada plants | 0.305 | 69 |
Cicada plants | 0.226 | 29 |
Cicada plants | 0.223 | 28 |
Cicada plants | 0.211 | 21 |
Cicada plants | 0.142 | 4 |
Cicada plants | 0.277 | 62.5 |
Cicada plants | 0.235 | 33 |
Cicada plants | 0.296 | 67.5 |
Cicada plants | 0.217 | 25 |
Cicada plants | 0.193 | 14 |
Cicada plants | 0.257 | 50.5 |
Cicada plants | 0.276 | 60.5 |
Cicada plants | 0.229 | 31 |
Control group | 0.212 | 22.5 |
Control group | 0.261 | 52.5 |
Control group | 0.203 | 18 |
Control group | 0.215 | 24 |
Control group | 0.178 | 9 |
Control group | 0.29 | 65 |
Control group | 0.268 | 59 |
Control group | 0.246 | 43 |
Control group | 0.241 | 39 |
Control group | 0.188 | 10 |
Control group | 0.265 | 56 |
Control group | 0.241 | 39 |
Control group | 0.285 | 64 |
Control group | 0.244 | 41 |
Control group | 0.253 | 47 |
Control group | 0.19 | 11.5 |
Control group | 0.145 | 5 |
Control group | 0.263 | 54.5 |
Control group | 0.135 | 2 |
Control group | 0.257 | 50.5 |
Control group | 0.198 | 16 |
Control group | 0.19 | 11.5 |
Control group | 0.249 | 45 |
Control group | 0.196 | 15 |
Control group | 0.247 | 44 |
Control group | 0.253 | 47 |
Control group | 0.17 | 7 |
Control group | 0.155 | 6 |
Control group | 0.266 | 57.5 |
Control group | 0.212 | 22.5 |
Control group | 0.253 | 47 |
Control group | 0.22 | 27 |
Control group | 0.14 | 3 |
n | sum of ranks | |
39 | 1567 | Cicada plants |
33 | 1061 | Control group |
72 | 2628 | total |
1423.50 | |
88.47 | standard deviation |
1.62 | z, corrected for ties |
.0530 | p-value (one-tailed, upper) |
As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,
Thus, we do not have the sufficient evidence to conclude that dead cicadas increase seed masses than the control group.
Want to see more full solutions like this?
Chapter 27 Solutions
Practice of Statistics in the Life Sciences
- Let X be a random variable with support SX = {−3, 0.5, 3, −2.5, 3.5}. Part ofits probability mass function (PMF) is given bypX(−3) = 0.15, pX(−2.5) = 0.3, pX(3) = 0.2, pX(3.5) = 0.15.(a) Find pX(0.5).(b) Find the cumulative distribution function (CDF), FX(x), of X.1(c) Sketch the graph of FX(x).arrow_forwardA well-known company predominantly makes flat pack furniture for students. Variability with the automated machinery means the wood components are cut with a standard deviation in length of 0.45 mm. After they are cut the components are measured. If their length is more than 1.2 mm from the required length, the components are rejected. a) Calculate the percentage of components that get rejected. b) In a manufacturing run of 1000 units, how many are expected to be rejected? c) The company wishes to install more accurate equipment in order to reduce the rejection rate by one-half, using the same ±1.2mm rejection criterion. Calculate the maximum acceptable standard deviation of the new process.arrow_forward5. Let X and Y be independent random variables and let the superscripts denote symmetrization (recall Sect. 3.6). Show that (X + Y) X+ys.arrow_forward
- 8. Suppose that the moments of the random variable X are constant, that is, suppose that EX" =c for all n ≥ 1, for some constant c. Find the distribution of X.arrow_forward9. The concentration function of a random variable X is defined as Qx(h) = sup P(x ≤ X ≤x+h), h>0. Show that, if X and Y are independent random variables, then Qx+y (h) min{Qx(h). Qr (h)).arrow_forward10. Prove that, if (t)=1+0(12) as asf->> O is a characteristic function, then p = 1.arrow_forward
- 9. The concentration function of a random variable X is defined as Qx(h) sup P(x ≤x≤x+h), h>0. (b) Is it true that Qx(ah) =aQx (h)?arrow_forward3. Let X1, X2,..., X, be independent, Exp(1)-distributed random variables, and set V₁₁ = max Xk and W₁ = X₁+x+x+ Isk≤narrow_forward7. Consider the function (t)=(1+|t|)e, ER. (a) Prove that is a characteristic function. (b) Prove that the corresponding distribution is absolutely continuous. (c) Prove, departing from itself, that the distribution has finite mean and variance. (d) Prove, without computation, that the mean equals 0. (e) Compute the density.arrow_forward
- 1. Show, by using characteristic, or moment generating functions, that if fx(x) = ½ex, -∞0 < x < ∞, then XY₁ - Y2, where Y₁ and Y2 are independent, exponentially distributed random variables.arrow_forward1. Show, by using characteristic, or moment generating functions, that if 1 fx(x): x) = ½exarrow_forward1990) 02-02 50% mesob berceus +7 What's the probability of getting more than 1 head on 10 flips of a fair coin?arrow_forward
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman