Find the 43rd term of the sequence -3.0,-2.5,-2.0,-1.5
1 answer:
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Answer:
Since the significance level is 0.05 we see that so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:
b. at least some, but not all, of the gestation periods across all four species are the same
Because is only to identify if AT LEAST one mean is different, NOT to conclude that the all the means are different.
Step-by-step explanation:
Previous concepts
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
Solution to the problem
The hypothesis for this case are:
Null hypothesis:
Alternative hypothesis: Not all the means are equal
In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.
If we assume that we have groups and on each group from
we have
individuals on each group we can define the following formulas of variation:
And we have this property
And in order to test this hypothesis we need to ue an F statistic and for this case the p value calculated is
Since the significance level is 0.05 we see that so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:
b. at least some, but not all, of the gestation periods across all four species are the same
Because is only to identify if AT LEAST one mean is different NOT to conclude that the all the means are different.
Answer:
95
Step-by-step explanation:
let x be his mark in the fourth test, then
mean = = 83 ( multiply both sides by 4 )
83 + 81 + 73 + x = 332
237 + x = 332 ( subtract 237 from both sides )
x = 95