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GalinKa [24]
3 years ago
7

Plz help,,,,,,,,,,,,,,,,,,,,

Mathematics
1 answer:
Kay [80]3 years ago
7 0
I believe that the answer would be c . 
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The percentage of adults who currently play chess (at least once during the past year) is 12% in the UK. Suppose that in a rando
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Answer:

p represent the population parameter , true proportion of people who play chess in the UK

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

z=\frac{0.1675 -0.12}{\sqrt{\frac{0.12(1-0.12)}{400}}}=2.923  

Step-by-step explanation:

Data given and notation

n=400 represent the random sample taken

X=67 represent the people who play chess

\hat p=\frac{67}{400}=0.1675 estimated proportion of people who play chess

p_o=0.12 is the value that we want to test

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

p represent the population parameter , true proportion of people who play chess in the UK

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion i higher than 0.13.:  

Null hypothesis:p \leq 0.12  

Alternative hypothesis:p > 0.12  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info required we can replace in formula (1) like this:  

z=\frac{0.1675 -0.12}{\sqrt{\frac{0.12(1-0.12)}{400}}}=2.923  

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