At 13% significance level, there isn't enough evidence to prove the administrators to claim that the mean score for the state's eighth graders on this exam is more than 280.
<h3>How to state hypothesis conclusion?</h3>
We are given;
Sample size; n = 78
population standard deviation σ = 37
Sample Mean; x' = 280
Population mean; μ = 287
The school administrator declares that mean score is more (bigger than) 280. Thus, the hypotheses is stated as;
Null hypothesis; H₀: μ > 280
Alternative hypothesis; Hₐ: μ < 280
This is a one tail test with significance level of α = 0.13
From online tables, the critical value at α = 0.13 is z(c) = -1.13
b) Formula for the test statistic is;
z = (x- μ)/(σ/√n)
z = ((280 - 287) *√78 )/37
z = -1.67
c) From online p-value from z-score calculator, we have;
P[ z > 280 ] = 0.048
d) The value for z = -1.67 is smaller than the critical value mentioned in problem statement z(c) = - 1.13 , the z(s) is in the rejection zone. Therefore we reject H₀
e) We conclude that at 13% significance level, there isn't enough evidence to prove the administrators to claim that the mean score for the state's eighth graders on this exam is more than 280.
Read more about Hypothesis Conclusion at; brainly.com/question/15980493
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Answer: The amount of carbon-14 remaining after 4 years is 99.95 grams.
Step-by-step explanation:
Hi, to answer this question we simply have to substitute t=4 in the equation given and solve for c.
c= 100 (0.99988)^t
c =100 (0.99988)^4
c = 100 x 0.999520086
c= 99.95200864 ≅99.95 grams (rounded)
The amount of carbon-14 remaining after 4 years is 99.95 grams.
Feel free to ask for more if needed or if you did not understand something.
Answer:
D) 1/36
Step-by-step explanation:
You are trying to roll a 3 the first time, and then a 5 the second time. This means that order matters.
It is given to you, that you are rolling a standard die (6-sides). The dice is ordered in the following: 1, 2, 3, 4, 5, 6
The first time, you are rolling for a 3. Note that there is only one 3 in the sequence, so your chance of rolling a 3 is 1 out of 6: 1/6.
The second time, you are rolling for a 5. Note that there is only one 5 in the sequence, so your chance of rolling a 5 is 1 out of 6: 1/6.
Next, multiply the two fractions together to find the probability of rolling a 3 and then a 5:
1/6 x 1/6 = (1 x 1)/(6 x 6) = 1/36
D) 1/36 is your answer.
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Answer:
D
Step-by-step explanation:
