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Lemur [1.5K]
3 years ago
14

A reporter for a large sports television network wants to determine if "homefield advantage" is dependent upon the sport. He col

lects data from a sample of games played in the four major professional sports in the country, and he records the number of home team wins and visiting team wins for these games. Use the 0.10 significance level to test the claim that home/visitor team wins are dependent on the sport. Find the value of the P-VALUE that would be used in a hypothesis test of the claim.
Mathematics
1 answer:
Ne4ueva [31]3 years ago
8 0

Answer:

with 0.10 level of significance the P-VALUE that would be used in the hypothesis claim is 0.05%

Step-by-step explanation:

In hypothesis testing in statistics, we can say that the p-value is a probability of obtaining test results when we assume that the null hypothesis is correct.

The p-value is the probability that the null hypothesis is true.

A p-value less than or equals to 0.05  is statistically significant. It shows strong evidence against the null hypothesis, meaning  there is less than a 5% probability the null is correct and clearly we can say that the results are random.

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B) The cost price of an article was Rs 4000.Find the marked price of the
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Market price = Rs. 6,000

Step-by-step explanation:

Given:

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7 0
3 years ago
Bill and George go target shooting together. Both shoot at atarget at the same time. Suppose Bill hits the target withprobabilit
german

Answer: (a) \frac{2}{9}       (b) \frac{6}{41}

Step-by-step explanation:

(a) P( Bill hitting the target) = 0.7        P( Bill not hitting the target) = 0.3

    P( George hitting the target) = 0.4     P(George not hitting the target) = 0.6

Now the chances that exactly one shot hit the target is = 0.7 x 0.6 + 0.4 x 0.3

                                                                                            = 0.54

Chances that George hit the target is = 0.4 x 0.3 = 0.12

So given that exactly one shot hit the target, probability that it was George's shot = \frac{0.12}{0.54} = \frac{2}{9} .

(b) The numerator in the second part would be the same as of (a) part which is 0.12.

The change in the denominator will be that now we know that the target is hit so now in denominator we include the chance of both hitting the target at same time that is 0.4 x 0.7 and the rest of the equation is same as above i.e.

Given that the target is hit,probability that George hit it =                                                 \frac{0.4\times 0.3}{0.7\times 0.6 +0.4\times 0.3+0.4\times 0.7}  = = \frac{6}{41}                                                                        

                                                                                           

6 0
3 years ago
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