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Citrus2011 [14]
3 years ago
9

Sociologists say the percentage of the classic music lovers is about 5%. From the survey, we know that only 83 out of 2000 peopl

e listen to the classic music. Let denote the proportion of the classic music lovers. The hypotheses are H0: p=0.05 vs Ha: p≠0.05. The p-value for this two-tailed test based on the normal distribution is 0.09. What is the p-value for a chi-square test?
Mathematics
1 answer:
disa [49]3 years ago
5 0

Answer:

p value from chi square = 0.081

Step-by-step explanation:

Given that sociologists say the percentage of the classic music lovers is about 5%. From the survey, we know that only 83 out of 2000 people listen to the classic music. Let denote the proportion of the classic music lovers. The hypotheses are H0: p=0.05 vs Ha: p≠0.05

p value = 0.09

Instead of this if we do chi square

H0: Proportion of music lovers is 5% and non music lovers is 95%

Ha: Proportions are not as per the estimate

(Two tailed chi square test)

Observed       83           1917       2000

Expected        100          1900     2000

O-E                  -17              17          0

chi square

(O-E)^2/E        2.89         0.1521     3.0421

df =1

p value from chi square = 0.081

Though p value is not exactly the same, the conclusion in both tests are the same

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The decision rule is  

Fail to reject the null hypothesis

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Step-by-step explanation:

From the question we are told that

  The population mean is  \mu = 15 \ months

   The sample size is  n =  25

   The sample mean is  \= x  = 17 \ months

   The standard deviation is  s = 5.5 \ months

Let assume the level of significance of this test is  \alpha  = 0.05

    The null hypothesis is  H_o :  \mu = 15

     The alternative hypothesis is  H_a : \mu  > 17

Generally the degree of freedom is mathematically represented as  

         df = n -1

=>       df = 25 -1

=>       df = 24

Generally the test statistics is mathematically represented as

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

=>   t = \frac{ 17 - 15 }{\frac{5.5}{\sqrt{25} } }

=>   t =  1.8182

Generally from the student t distribution table the probability of obtaining   t =  1.8182 to the right of the curve at a degree of freedom of df = 24  is  

    p-value  = P(t > 0.18182 ) = 0.4286

From the value obtained we see that  p-value > \alpha hence

The decision rule is  

Fail to reject the null hypothesis

 The conclusion is  

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Step-by-step explanation:

.......................

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Answer:

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Step-by-step explanation:

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