The results of the t-test for zinnia height: p=0.381.Did these zinnia height results support the alternative hypothesis of this study?Yes, the zinnia were taller in the mycorrhizae soitNo, there was no statistical difference in zinnia heights between the soils.Question 14✓ SavedThe results of the t-test for zinnia weight: p=0.217.For zinnia weight, can we reject the null hypothesis of the study?Yes, zinnia weighed more in the mycorrhizae soilNo. We failed to reject the null hypothesis because no statistical difference wasfound in zinnia weight between the two soil types

The t test is used when we have to check significant difference between two meaningsand p value is…

Answered: The results of the t-test for zinnia…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Your company is evaluating two cloud-based secured data storage services. "Great Sky," the newer service, claims its uploading and downloading speeds are faster than the older service, "Clear and Steady Skies." You need to make a decision based on published access times for both services at different times and for varying file sizes. To make your decision, you purchase two statistical studies from which you discover the average download time for Great Sky is 0.67 sec. per MB and for Clear Steady Skies is 0.84. Additionally, the studies indicate the following sample information: NGS = 60 ncss = 50, SGS = 0.25 and scss = 0.55. Use a .05, and the five step hypothesis testing procedure to make the appropriate decision. Prepare an Excel worksheet properly labeled and well formatted to share with your boss (a very picky person that truly appreciates good looking work product).
The following data were obtained from an independent-measures study comparing three treatment conditions. Treatment 1 2 3 n=6 n=6 n=6 M=1 M=2 M=6 SS=60 SS=65 SS=40 N= 18 G= 54 Sum of squares X^2= 411 Conduct an ANOVA with to determine whether there are any significant differences among the three treatment means. (alpha=.05)
How do you evaluate the coefficients for the independent variables against the size and statistical significance of the coefficient for the location variable to reject or fail to reject the hypothesis?.
Hello, Will you help me solve this problem and explain on a ti 84? Thank you. I am confused on which lists to use for L1 and L2. ex: You wish to test the following claim (HaHa) at a significance level of α=0.01α=0.01. Ho:μ1=μ2Ho:μ1=μ2 Ha:μ1≠μ2Ha:μ1≠μ2You obtain the following two samples of data. Sample #1 Sample #2 68.2 65.4 62.7 72.1 67.6 72.7 65.3 77.6 54.4 59.1 70.4 67.2 73.5 69.2 64.9 72.7 68.5 56.6 57.5 58.8 65 70.2 67.5 66.6 64.9 67.6 65.4 59.6 69.2 66.1 63.3 66.1 73.1 72.4 63 63.5 59.6 61.2 61 59.8 61.7 64.9 64.1 67.9 67.3 77.6 64.5 82.9 65.4 68.9 69.9 70.3 57.8 62.4 74 49.8 60.7 90.4 79.6 60.3 79.2 78.8 80.9 80.5 76.1 77.6 48.6 77.2 68.2 83.4 94.1 77.6 83.9 71.6 43.1 83.4 61.6 59.8 94.1 54.9 90.4 98.1 What is the test statistic for this sample? (Report answer accurate to three decimal places.)test statistic = What is the p-value for this sample? For this calculation, use the degrees of freedom…
Classify the cases/scenario as regards to the appropriate design/analysis for the experiments: A woman learned from a friend that rubbing the leaves with either egg white or coconut oil could enhance the appearance of indoor plants. But she was doubtful that this will induce withering of the leaves. She then plans to conduct an experiment to compare the effects of these two. For better comparison, she would like to include a control (no egg whole nor coconut oil) and the outcome data follows a non-normal distribution. * (Parametric/Nonparametric – Design/Analysis e.g. Par-ANOVA(RCBD), Nonpar-Wilcoxon test) Your answer
Analyze data in SPSS using a dependent samples t-test • Imagine that we are interested in the difference in heart rate pre and post doing 30 seconds of jumping jacks. How might we test this question? (basic research design) Do the 4 steps of hypothesis testing (use the SPSS output sent): Step 1: • Write null hypothesis • Write research hypothesis Step 2:• Use a two-tailed test with alpha set to  = .05** • what is the critical region? Step 3: • what is the p-value? • what is the t-statistic? Step 4: • determine whether results are statistically significant • what do you conclude in general?
Look at the attached images. There is the normality test and the Mann Whitney test for the two groups of INR and PCV (PCV1/a, PCV2/b , INR1/a, INR2/b). Interpret the data using both tests and explain why there may or may not be differences between the two cohorts.
Considering that k= 4, n= 58, SSE= 84.75, and SSR= 725 fill in the ANOVA table and state the results of testing the stated Null Hypothesis below DF SS MS Significance Regression 0.455 Residual From the table F= 2.55 Total Họ. by = 0 the result of testing the Null Hypothesis is
A researcher notes that there seems to be a difference in the prevalence of high blood pressure among college-educated individuals who consume low amounts of processed foods and the prevalence of individuals who only have a high school education and consume high amounts of processed foods. Use the appropriate hypothesis to test for the independence of the two independent variables presented here at the 5% significance level to ensure confounding has not influenced the study’s results. Then, interpret your response. Diet Low in Processed Foods Normal BP High BP Total College Education 124 55 179 High School Education 69 152 221 Total 193 207 400 Diet High in Processed Foods Normal BP High BP Total College Education 64 85 149 High School Education 98 153 251 Total 162 238 400 The chi square value of 37.724 is higher than the chi square value of 3.84; therefore, we can accept H0 at…
my answer for part b) of 6.24 was: (4.3, 9.26) so the variability in Hours is equal to 4.3 hours to 9.26 hours and in minutes its 258 to 555.6 minutes. I also posted 6.24 the previous exercise so that you have all the information but I need help with 6.25.
My answer is as follows: T critical = 2.70 T value= 7.775 P< a (0.01) Therefore we reject the null. Is this correct?
Which combination would lead you to Fail to Reject H0? ANOVA: Fobt = 3.61 Fcrit = 2.55 2-tailed t-test: tobt = 2.36 tcrit = 1.89 Correlation: robt = 0.68 rcrit = 0.64 Chi-Square GOF: χ2obt = 3.24 χ2crit = 4.25

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