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Delvig [45]
2 years ago
11

A statistician believes that women received more bachelor's degrees than men did last year. To test this claim, she selects coll

eges and universities randomly and compares the number of bachelor's degrees awarded to men and women. Suppose that data were collected for a random sample of 7 colleges and universities, where each difference is calculated by subtracting the number of degrees earned by men from the number of degrees earned by women. Assume that the number of degrees is normally distributed. The statistician uses the alternative hypothesis Ha:μd>0. Using a test statistic of t≈2.968, which has 6 degrees of freedom, determine the range that contains the p-value.
Mathematics
1 answer:
pychu [463]2 years ago
3 0

Answer:

0.0125

Step-by-step explanation:

The P-value can be calculated by using excel function T.DIST.RT(2.968,6).

Right tail is used in the function because our alternative hypothesis has greater than sign (>) whereas 2.968 represent test statistic value and 6 represents degree of freedom. The resultant p-value from excel function is 0.0125.

Or if we use t- distribution right tail area table, we see that 6 for degree of freedom the value in table is there are two value 2.44691 and 3.14627 which corresponds to p=0.025 and p=0.01. As 2.44691 is smaller than 2.968 so we take p=0.01 against 3.14627. Thus, in this scenario p-value would be 0.01.

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Blababa [14]

Given:

A bowl contains 25 chips numbered 1 to 25.

A chip is drawn randomly from the bowl.

To find:

The probability that it is

a. 9 or 10?

b. even or divisible by 3?

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a. We have,

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Favorable out comes are either 9 or 10. So,

Number of favorable outcomes = 2

The probability that the selected chip is either 9 or 10 is:

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b. The numbers that are even from 1 to 25 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.

The numbers from 1 to 25 that are divisible by 3 are 3, 6, 9, 12, 15, 18, 21, 24.

The numbers that are either even or divisible by 3 are 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24.

Number of favorable outcomes = 16

The probability that the selected chip is either even or divisible by 3 is:

\text{Probability}=\dfrac{16}{25}

Therefore, the probability that the selected chip is either even or divisible by 3 is \dfrac{16}{25}.

c. The numbers from 1 to 25 that are divisible by 5 are 5, 10, 15, 20, 25.

The numbers from 1 to 25 that are divisible by 10 are 10, 20.

The numbers that are divisible by both 5 and 10 are 10 and 20.

Number of favorable outcomes = 2

The probability that the selected chip is divisible by 5 and divisible by 10 is:

\text{Probability}=\dfrac{2}{25}

Therefore, the probability that the selected chip is divisible by 5 and divisible by 10 is \dfrac{2}{25}.

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