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OverLord2011 [107]
3 years ago
9

PLEAAAAAASEEEE HELPPPP ASAAPPPP 95 POINTS JUST FOR THIS

Mathematics
1 answer:
zloy xaker [14]3 years ago
5 0

Answer:

see below

Step-by-step explanation:

123,116,109,102,95.

First find the common difference

d = 116 - 123 = -7

We are subtracting 7 each time

Using the formula

a1 = 123

d=-7

an = 123+ (n-1)(-7)

We need to find the 100th term

Let n = 100

a100 = 123 +(100-1) (-7)

         = 123+(99)(-7)

        = 123-693

        = -570

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Answer:

t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551    

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Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

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Step-by-step explanation:

Information provided

\bar X=12.2 represent the sample mean fould against

s=1.6 represent the sample standard deviation

n=34 sample size  

represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

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We need to conduct a hypothesis in order to check if the true mean is higher than 11.5 fouls per game:  

Null hypothesis:\mu \leq 11.5  

Alternative hypothesis:\mu > 11.5  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

The statistic is given by:

t=\frac{12.2-11.5}{\frac{1.6}{\sqrt{34}}}=2.551    

P value

The degreed of freedom are given by:

df = n-1=34-1=33

Since is a one side test the p value would be:  

p_v =P(t_{(33)}>2.551)=0.008  

Since the p value is less than the significance level of 0.05 we have enough evidence to reject the null hypothesis in favor of the claim

And the best conclusion for this case would be:

b)The p-value is 0.008, indicating sufficient evidence for his claim.

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