Answer:
The variance in weight is statistically the same among Javier's and Linda's rats
The null hypothesis will be accepted because the P-value (0.53 ) > ∝ ( level of significance )
Step-by-step explanation:
considering the null hypothesis that there is no difference between the weights of the rats, we will test the weight gain of the rats at 10% significance level with the use of Ti-83 calculator
The results from the One- way ANOVA ( Numerator )
with the use of Ti-83 calculator
F = .66853
p = .53054
Factor
df = 2 ( degree of freedom )
SS = 23.212
MS = 11.606
Results from One-way Anova ( denominator )
Ms = 11.606
Error
df = 12 ( degree of freedom )
SS = 208.324
MS = 17.3603
Sxp = 4.16657
where : test statistic = 0.6685
p-value = 0.53
level of significance ( ∝ ) = 0.10
The null hypothesis will be accepted because the P-value (0.53 ) > ∝
where Null hypothesis H0 = ∪1 = ∪2 = ∪3
hence The variance in weight is statistically the same among Javier's and Linda's rats
I’m just going off of what I was told in fifth grade, and we were told it was finger nails. Hope this helps!
Answer:



Step-by-step explanation:
Given


Solving (a): The mean
Mean is calculated as:

This gives:



Solving (b): The median
Sort the data in ascending order:


The median is:



The 8th item on the sorted dataset is 7; So:

Solving (c): The mode

Because it has a frequency of 3 (more than any other element of the dataset).