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Sergeeva-Olga [200]
3 years ago
5

When using a one-tailed test and a sample size of 24, what is the probability of a value being to the left of a number that has

a student-t statistic of –2.50?
Mathematics
1 answer:
UNO [17]3 years ago
3 0

Answer:

P-value = 0.009997                                                  

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 24

Nature of test: One-tailed test(Left tailed test)

t_{\text{statistic}} = -2.50

We have to calculate the p-value.

Degree of freedom = n - 1 = 23

We calculate the p-value from the table corresponding to t score -2.50 and degree of freedom 23.

P-value = 0.009997

0.009997 is the  probability of a value being to the left of a number that has a student-t statistic of –2.50.

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Rewrite the expression in the form y^n
bazaltina [42]

Answer:

$ y^6 $

Step-by-step explanation:

$ \frac{a^b}{a^c} = a^{b - c} $

Using this rule we have:

$ \frac{y^{-7}}{y^{-13}}  = y^{-7 -(-13)} = y^6$

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Andrews [41]
The correct option is: Option (C) 2

Explanation:

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3 years ago
I REALLY NEED HELP! PLEASE HELP ME!!
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5:3

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2 years ago
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Lana71 [14]

The answer would be 14/5.

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So  1 /7 × 5/ 5 = 5 /3/5

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Now we have  5 /35  and  14 /35

Then all you have to do is divide Joe's by Jane's to get the answer.

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Meaning the answer is 14/5.

Hope I could help! :)


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