Question

having trouble working excel, please help

Lab Assignment Each student has been assigned a unique data set, and will be required to perform basic statistical analyses on the data. Each exercise should be saved to a new file so one of your first tasks will be to make sufficient copies of your data file, named appropriately. E.g. Grant, Karen Inf Stats Ex1; Grant, Karen Inf Stats Ex2; Grant, Karen Inf Stats Ex3. All data files are taken from real data collected on raising tadpoles of Rana sylvatica. Some of the data sets use snout-vent length (SVL) and are measured in mm. The other data sets use mass as the measurement and the units are grams. Exercise 1. Determining normality of data: perform the descriptive statistics and histogram (with chart output) analyses on your data on each of the three groups. Output for the descriptive statistics and histogram for each treatment group should be posted to a new sheet (control analyses to sheet #2, treatment 1 to sheet #3, treatment 2 to sheet #4). Label each sheet tab appropriately, insert a header with your name and a footer with a description of each sheet. Exercise 2. Normally, when comparing control and treatment groups pair-wise analyses are not pe using Analysis of Variance (ANOVA). Your data is single factored (light treatment) so you will be using a single factor ANOVA rather than a double factor rformed but rather all groups are compared at the same time or repeated measures ANOVA. As always, put your output for the analysis on a new sheet, name the sheet, name in header, description of sheet in footer Exercise 3. One of the main reasons for performing an experiment is to determine if there is a difference in a variable in control versus treatment groups. Ultimately you will be comparing the means of the 3 possible pairs (Control versus Treatment 1, Control versus Treatment 2, & Treatment 1 versus Treatment 2). For all of these, the null hypothesis would be mean1 mean2. Due Mann-Whitney are not available so we are going to have to use parametric tests If you click on the data analysis tool and scroll through the list of options, you to limitations of the statistics package in Excel, nonparametric tests such as (assumes normal distribution). will note that there are several options possible when using t-tests (but not all appropriate). The first option is for a "T-test paired sample'. This would be sed in a before and after scenario which is not what we have in our data sets e are comparing two data sets (containing different individuals) but there may wrinces hetween these two (which we must first Inferential Statistics Name: Descriptive Statistics (Exercise 1) Treatment2 Treatment1 Control Mean Variance Kurtosis Skewness Normal/Non-normal distribution ased upon the above chart and the three chart outputs, would you say that the data in each group is normal or non-normal in distribution? Explain. Analysis of variance -single factor (Exercise 2) F critical (Accept/reject null hypothesis Dfo 0.05 According to the ANOVA, is there a difference between control and treatment groups? Explain. Does this test tell you where the differences have occurred? Statistical analyses available through the EXCEL software are extremely limited compared with programs such as SPSS (which can perform all analyses outlined in Chapter 10 of your lab text). Had SPSS been available, would you have analyzed your data using different tests? Explain. (N.B. another test available through SPSS would have been post hoc tests (Scheffe's, Tukey's, Bonferonni, LSD). Used with ANOVAs they show where the differences between test groups lie) Inferential Statistics Name: F-Test Two Sample for Variances (Exercise 3) Control vs Treatment 1Control vs Treatment 2 Treatment 1 vs Treatment2 Variable 1 Variable 2Variable 1 Variable 2 Variable 1 Variable 2 Variance df F critical two- tailed a- 0.05 Appendix 8 Null hypothesis accepted/rejected T-test to be used T-test Summaries Control vs Treatment 1 Control vs Treatment 2 Treatment 1 vs Treatment2 Control Treatment 1 Control Treatment 2 Treatment 1 Treatment 2 Treatment Mean df T stat T critical two- tailed a 0.05 Null hypothesis accepted/rejected Based on the t-tests, is there a significant difference between any of the groups? Explain. Make sure that you submit the summary sheets on paper and the Excel files electronically Page 2 of 2 Control Treatment 1 Treatment 2 0 6.5 10.0 10.0 11.0 9.5 8.5 8.0 9.0 8.0 10.0 10.0 10.5 11.0 10.0 12.0 12.0 8.5 8.0 9.5 10.0 11.0 11.0 9.0 9.0 8.0 7.0 9.0 8.0 8.0 8.0 9.0 8.5 9.0 10.0 8.0 9.0 9.0 9.0 8.0 10.0 9.5 9.5 8.0 10.0 9.0 10.0 10.0 10.0 10.0 9.5 10.0 9.0 19.0 10.0 10.0 10.0 8.5 9.0 9.0 9.5 9.0 9.0 9.0 8.0 8.0 8.5 9.0 8.5 7.5 8.0 An 8.0 8.5 9.5 11.0 10.0 9.0 9.0 9.0 8.0 11.5 10.0 12.5 9.0 9.0 10.5 8.0 9.0 Accor Explair 9.0 9.0 9.0 9.5 10.0 11.0 8.0 11.0 11.0 9.5 8.0 9.5 .O 00 .5 .0 5 0 0 5 .0 .0 0 5 0 0 0 0 0 .0 .0 5050000 898 10 001-9. 10 10 9 9 9 900 00000 .0 0 .0 5 0 0 0 0 0 0 00 .0 5 8 0 0 0 0 5 0 0 .0 0 8 99798 10 9。9 2 11188 8. 0 0 08 0 1979 05 .0 0 998 10 10 998 100-9. 0 1 10 8 1 1 1 1

Answer 1

	CONTROL	TREATMENT 1	TREATMENT 2
MEAN	9.564	9.047	9.25
VARIANCE	2.113	1.15	1.063
KURTOSIS	23.76	3.971	2.319
SKEWNESS	3.454	0.7914	0.1874
NORMAL/NON NORMAL	NON NORMAL	CLOSE TO NORMAL	CLOSE TO NORMAL

control treatment 1 treatment 2

Mean:	Mean:	Mean:
9.564	9.047	9.25
Median:	Median:	Median:
9.5	9	9
Mode:	Mode:	Mode:
10	9	9

the control variable is not normally distributed because mean, median and the mode is not the same and skewness value is so high(3.454). the treatment variable is closely normally distributed because mean, median, and the mode is closely same and skewness value is low(0.7914). the treatment2 is most closely normally distributed because of mean. median and mode is closely same and skewness value is close to zero(0.1874)

	CONTROL & TREATMENT 1		CONTROL & TREATMENT 2		TREATMENT 1 & TREATMENT 2
	VARIABLE 1	VARIABLE 2	VARIABLE 1	VARIABLE 2	VARIABLE 1	VARIABLE 2
VARIANCE	2.113	1.15	2.113	1.063	1.15	1.063
DF	77	74	77	73	74	73
F	3.376		3.951		1.17
CRITICAL VALUE AT 0.05	1.465		1.467		1.47
NULL HYPOTHESIS REJECTED/ ACCEPTED	REJECTED		REJECTED		ACCEPTD
T-TEST TO BE USED	YES		YES		YES

according to ANOVA ( F ratio), there is a significant difference between control and treatmenat1 because the calculated F value is greater than the critical value. also, there is a significant difference between control and treatment 2 because the calculated F value is greater than the critical value.

Based on the F test we can say that there is a difference between the variance of control &  treatment 1 and control & treatment 2.

if SPSS available please use SPSS software because it is very easy and time conserving software. we can perform any statistical analysis and test by giving only his variable then the software automatically calculates which are you want.

	CONTROL & TREATMENT 1		CONTROL & TREATMENT 2		TREATMENT 1 & TREATMENT 2
	VARIABLE 1	VARIABLE 2	VARIABLE 1	VARIABLE 2	VARIABLE 1	VARIABLE 2
MEAN	9.564	9.047	9.564	9.25	9.047	9.25
DF	151		150		147
T-TEST	2.4956		1.523		-1.17
CRITICAL VALUE AT 0.05	1.655		1.655		1.655
NULL HYPOTHESIS REJECTED/ ACCEPTED	REJECTED		ACCEPTED		ACCEPTD

Based on t-test there is a significant difference between control and treatment 1 because calculated t value is greater than the critical value. so null hypothesis rejected and alternative hypothesis accepted.