When I did this problem I got a very small P value and was told it is wrong, I do not know which other test would be appropriate for this question.
Here is a data set containing plant growth measurements of plants grown in solutions of commonly-found chemicals in roadway runoff.
Distilled H2O | Petro | NaCl | MgCl | NaCl + MgCl |
19.93 | 19.85 | 19.87 | 19.91 | 19.73 |
19.91 | 20.06 | 19.88 | 19.92 | 19.77 |
20.08 | 19.99 | 20.04 | 19.84 | 19.75 |
19.99 | 19.88 | 20.05 | 19.98 | 19.93 |
19.9 | 19.98 | 20.06 | 19.82 | 19.94 |
19.98 | 20.08 | 19.83 | 19.92 | 19.79 |
19.92 | 20.1 | 19.9 | 20.09 | 19.84 |
20.01 | 19.82 | 19.83 | 20.1 | 19.94 |
19.96 | 20.01 | 19.85 | 20.04 | 19.89 |
20.13 | 20.1 | 19.87 | 20.04 | 19.72 |
20.15 | 19.84 | 19.85 | 19.87 | 19.88 |
20.04 | 20.03 | 19.93 | 19.89 | 20 |
19.98 | 20.01 | 19.82 | 19.77 | 19.74 |
20.03 | 19.96 | 19.85 | 19.97 | 19.95 |
20.13 | 19.91 | 20.06 | 19.84 | 19.79 |
20 | 20.03 | 20.04 | 20.07 | 19.85 |
20.07 | 19.92 | 20 | 19.83 | 19.74 |
19.98 | 19.94 | 19.9 | 19.9 | 19.78 |
20.02 | 20.01 | 19.94 | 19.95 | 19.88 |
19.94 | 19.8 | 20.05 | 19.78 | 19.83 |
Phragmites australis, a fast-growing non-native grass
common to roadsides and disturbed wetlands of Tidewater Virginia,
was grown in a greenhouse and watered with either:
Distilled water (control);
A weak petroleum solution (representing standard
roadway runoff);
Sodium chloride solution;
Magnesium chloride solution;
De-icing brine (50% sodium chloride and 50%
magnesium chloride).
Twenty grass preparations were used for each solution, and total
growth (in cm) was recorded after watering every other day for 40
days.
-Perform the correct statistical test to determine the
p-value.
-Report your answer rounded to four decimal places using scientific
notation (ex: 1.2345E-08)
-You should use formulas, functions, and the Data Analysis ToolPak
in MS Excel to avoid additive rounding errors. Here are some useful
functions:
=t.test(array1,array2,tails,type)
Produces a p-value for any type of a t-test from
a data set
=chidist(x,degrees of freedom)
Produces a p-value from a calculated chi-squared
statistic
ANOVA Single Factor from the data analysis tool pack
Performs a 1-way ANOVA on a data set with
only one factor
ANOVA 2-Factor from the data analysis tool pack
Performs an
ANOVA on data with multiple treatments
Regression from the data analysis tool pack
Performs a linear regression and the corresponding
t-test on the slope of the line
___________? (P value)
2) Which of the following is a correct statement of the alternate hypothesis?
Mean growth in the control group is no greater than mean growth in the experimental groups. |
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Mean growth in the NaCl + MgCl group is less than mean growth in the control group. |
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Mean growth in the control group is greater than mean growth in the experimental treatments. |
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Mean growth in one or more of the groups is greater than or less than one of the others. 3) Based on the p-value resulting from your analysis of the Phragmites dataset, which of the following is / are true? (select all that apply)
|
Seeinf the results of the t tests we conclude
Mean growth in the NaCl + MgCl group is less than mean growth in the control group.
See the left lower part of the second image for the test and upper right part to reject the option a
Mean growth in one or more of the groups is greater than or less than one of the others.
3) Based on the p-value resulting from your analysis of the Phragmites dataset, which of the following is / are true? (select all that apply)
Applied part is The p-value is less than 0.05.
The calculated F value was greater than the critical F value.
4) Which is / are (a) proper interpretation(s) of your p-value from your Phragmites data analysis? (select all that apply)
I would reject the null hypothesis.
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