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8_murik_8 [283]
4 years ago
12

Katy's measured the growth of a plant seedling. The seedling grew 1/3 inch by the end of the first week. The seedling grew 5/6 i

nch by the end of the second week. About how much did the seedling grow in the fist 2 weeks?
Mathematics
2 answers:
Vikki [24]4 years ago
8 0
In the given problem it is already given that the seedling grew by 1/3 inch in the first week and 5/6 inch in the second week. Then it would not be a problem to find the total length of growth of the seedling.
The length up to which the seedling grew in 2 weeks = (1/3) + (5/6)
                                                                                     = (2 + 5)/6
                                                                                     = 7/6 inch
So the seedling grew by 7/6 inches in two weeks. I hope that this answer is what you were looking for. You can solve such problems in the future by following the method that is applied for solving this problem.
ddd [48]4 years ago
5 0
1/3 + 5/6 = 2/6 + 5/6 = 7/6
It grew by 7/6 or 1 1/6 inches by the first 2 weeks.
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Answer:

II. This finding is significant for a two-tailed test at .01.

III. This finding is significant for a one-tailed test at .01.

d. II and III only

Step-by-step explanation:

1) Data given and notation    

\bar X=19.2 represent the battery life sample mean    

\sigma=2.5 represent the population standard deviation    

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\alpha represent the significance level for the hypothesis test.    

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We need to conduct a hypothesis in order to check if the mean battery life is equal to 18 or not for parta I and II:    

Null hypothesis:\mu = 18    

Alternative hypothesis:\mu \neq 18    

And for part III we have a one tailed test with the following hypothesis:

Null hypothesis:\mu \leq 18    

Alternative hypothesis:\mu > 18  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:    

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)    

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

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First we need to calculate the degrees of freedom given by:  

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p_v =2*P(t_{(24)}>2.4)=0.0245

And for part III since we have a one right tailed test the p value is:

p_v =P(t_{(24)}>2.4)=0.0122

5) Conclusion    

I. This finding is significant for a two-tailed test at .05.

Since the p_v. We reject the null hypothesis so we don't have a significant result. FALSE

II. This finding is significant for a two-tailed test at .01.

Since the p_v >\alpha. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

III. This finding is significant for a one-tailed test at .01.

Since the p_v >\alpha. We FAIL to reject the null hypothesis so we have a significant result. TRUE.

So then the correct options is:

d. II and III only

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Answer:

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