Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat's weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of 10%, test the hypothesis that the three formulas produce the same mean weight gain. Linda's rats Tuan's rats Javier's rats
43.5 47.0 51.2
39.4 40.5 40.9
41.3 38.9 37.9
46.0 46.3 45.0
38.2 44.2 48.6
Determine whether or not the variance in weight gain is statistically the same among Javier’s and Linda’s rats. Test at a significance level of 10%.

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 am currently in the exact same situation rn. algebra2/adv algebra is killing

step-by-step explanation:

percentage - 25%

probability - 0.25

we know,

total amount of students in class = 20

total boys = 10

total girls = 10

no of girls who wear glasses = 10/2 = 5

probability of choosing a girl with glasses is =

number of girls wearing glasses/total number of students in the class = 5/20

thus, percentage of selecting a girl wearing glasses will be

= 5/20 × 100

= 5 × 5

25%

therefore,

probability will be 0.25

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