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Thepotemich [5.8K]
3 years ago
9

A professor thinks that the students in her statistics class this term are less creative than most students at this university.

A previous study found that students at this university had a mean score of 35 on a standard creativity test and scores were normally distributed. The professor finds that her class of 23 students scores a mean of 33 on this scale, with a standard deviation of 5. Using a .01 significance level, what is the correct decision and conclusion
Mathematics
1 answer:
choli [55]3 years ago
3 0

Answer:

It is not enough evidence to reject null hypothesis. It is not enough evidence to say that mean score is not equal to 35.

Step-by-step explanation:

\mu=35\\n=23\\\=x=33\\s=5\\\alpha=0.1

1. Null and alternative hypothesis

H_{0}:\mu=35\\H_{1}:\mu

2. Significance level

\alpha=0.1\\1-\alpha=0.99

Freedom degrees is given by:

v=n-1\\v=23-1=22

For a sgnificance level of 0,01 and 22 freedom degrees,  t-student distribution value is:

t_{0.01;23}=-2. 5083

3. Test statistic

t=\frac{\=x-\mu}{\frac{s}{\sqrt{n} } }

t=\frac{33-35}{\frac{5}{\sqrt{23} } }

t=-1,918

In this case, we have an one left tailed analysis, it means that null hypothesis is rejected if t

-1.918>-2.5083

Conclusion:

It is not enough evidence to reject null hypothesis. It is not enough evidence to say that mean score is not equal to 35.

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0/1 For positive integer n, n? = n! · (n − 1)! · … · 1! And n# = n? · (n − 1)? · … · 1?. What is the value of 4# · 3# · 2# · 1#?
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Answer:

331776

Step-by-step explanation:

Since n? = n! · (n − 1)! · … · 1! And n# = n? · (n − 1)? · … · 1?

Then 4# = 4? · (4 − 1)? · (4 − 2)?· 1?

= 4? · 3? · 2?· 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 4? = 4! · (4 − 1)! · (4 − 2)! · 1! = 4! · 3! · 2! · 1! = 288

Thus, 3? = 3! · (3 − 1)! · 1! = 3! · 2! · 1! = 12

Also, 2? = 2! · (2 − 1)! · 1! = 2! · 1! · 1! = 2

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So, 4# = 4? · 3? · 2?· 1? = 288 × 12 × 2 × 1 = 6912

We now find 3#

3# = 3? · (3 − 1)? · 1? = 3? · 2?· 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 3? = 3! · (3 − 1)! · 1! = 3! · 2! · 1! = 12

Thus, 2? = 2! · (2 − 1)! · 1! = 2! · 1! · 1! = 2

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So, 3# = 3? · 2?· 1? = 12 × 2 × 1 = 24

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2# = 2? · (2 − 1)? · 1? = 2? · 1?· 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 2? = 2! · (2 − 1)! · 1! = 2! · 1! · 1! = 2

1? = 1! · (1 − 1)! · 1! = 1! · 0! · 1! = 1

So, 2# = 2?· 1? = 2 × 1 = 2

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1# = 1? · 1? = 1? · 1?

Now, n? = n! · (n − 1)! · … · 1!

So, 1? = 1! · (1 − 1)! · 1! = 1! · 0! · 1! = 1

And, 1# = 1? · 1? = 1 × 1 = 1

So,  4# · 3# · 2# · 1#? =  6912 · 24 · 2 · 1? = 331776

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