Jamal is buying ingredients to make a large batch of granola to sell at a school fair. He buys 3.8 pounds of walnuts for $4.61 p
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Answer:
The approximate difference in the half-lives of the isotopes is 66 days.
Step-by-step explanation:
The decay of an isotope is represented by the following differential equation:
Where:
- Current mass of the isotope, measured in kilograms.
- Time, measured in days.
- Time constant, measured in days.
The solution of the differential equation is:
Where is the initial mass of the isotope, measure in kilograms.
Now, the time constant is cleared:
The half-life of a isotope () as a function of time constant is:
The half-life difference between isotope B and isotope A is:
If ,
and
, the difference in the half-lives of the isotopes is:
The approximate difference in the half-lives of the isotopes is 66 days.
Answer:
a. two-tailed test
b. reject H0 if Z>1.96 or Z<-1.96
c. Z = 1.73
d. we have failed to reject the null hypothesis
e. P-value = 0.0418
Step-by-step explanation:
We will use a Z test to resolve this, an it will be a two-tailed test because the hypothesis statements are not indicating a specific direction for the significant difference (H0 : μ1 = μ2 ; H1 : μ1 ≠ μ2), this also means that the significanclevel will be divided between the both tails (2.5% en each tail for the rejection regions). See attached drawing for reference.
We need to find our critical value:
Zα/2 = Z(0.05/2) = 0.025
If we look for 0.025 in a Z table we will find that the critical value is 1.96 to the right, and by symmetry -1.96 to the left. So our decision rule will be to reject H0 if Z>1.96 or Z<-1.96
The Z test will be done using the next equation:
Z = (x⁻1 - x⁻2) - (μ1 - μ2) / √( σ²1/n1 + σ²2/n2
Because we are testing the null hypothesis we know that μ1 - μ2 must be zero if they are supposed to be equal (H0 : μ1 = μ2), so we calculate as follows:
Z = (100.5 - 98.8) - (0) / √( (3.4)²/38 + (5.8)²/51 = 1.7/0.9817 = 1.73
Z<1.96, therefore we have failed to reject the null hypothesis
The P-value for this test would be represented by the probability of Z being greater than 1.73, so we can look for it in any Z table, finding that its value is 0.0418, and because P-value>0.025 we again confirm that we don't have evidence statistically significant to reject the null hypothesis
