Answer:
Answer:
Since the calculated value of t= -1.340 does not fall in the critical region , so we accept H0 and may conclude that the data do not provide sufficient evidence to indicate hat there is difference in mean carbohydrate content between "meals with potatoes" and "meals with no potatoes".
Step-by-step explanation:
Potatoes : No Potatoes : Difference Difference (d)²
(Potatoes- No Potatoes)
29 41 -12 144
25 41 -16 256
17 37 -20 400
36 29 -7 49
41 30 11 121
25 38 -13 169
32 39 -7 49
29 10 19 361
38 29 9 81
34 55 -21 441
24 29 -5 25
27 27 0 0
<u>29 31 -2 4 </u>
<u> ∑ -64 2100 </u>
- We state our null and alternative hypotheses as
H0 : μd= 0 and Ha: μd≠0
2. The significance level alpha is set at α = 0.01
3. The test statistic under H0 is
t= d`/sd/√n
which has t distribution with n-1 degrees of freedom.
4. The critical region is t > t (0.005,12) = 3.055
5. Computations
d`= ∑d/n = -64/ 13= -4.923
sd²= ∑(di-d`)²/ n-1 = 1/n01 [ ∑di² - (∑di)²/n]
= 1/12 [2100- ( -4.923)] = 175.410
sd= √175.410 = 13.244
t = d`/sd/√n= - 4.923/13.244/√13
t= - 4.923/3.67344
t= -1.340
6. Conclusion :
Since the calculated value of t= -1.340 does not fall in the critical region , so we accept H0 and may conclude that the data do not provide sufficient evidence to indicate hat there is difference in mean carbohydrate content between "meals with potatoes" and "meals with no potatoes".
Answer:
3⁷
Step-by-step explanation:
it's kinda messy but my step by step is in the picture. practically when you "multiply" an exponent (as long as they have the same base number) with another like 3⁴×3⁵ or (3⁴)⁵ you add the exponents together so you'd get 3⁴×3⁵=3⁹
when dividing an exponent by another exponent (with again, the same base number) you subtract the exponents like this: 3⁶÷3² or 3⁶/3² would equal 3⁴
Answer:
Step-by-step explanation:
Given that three positive numbers have sum 18.
Let the numbers be 
Then product
To find maxima, let us use partial derivaties
Equate I derivatives to 0
Solving the two linear equations we get solution as
(6,6)
Hence maximum when x =y=z=6
i.e. when all numbers are equal to 6.
