How do you generalise that the probability of “head” occurring on a coin is 0.5?

$ \begingroup $ In a aboveboard 10 day experiment with 1000 tosses each day with a fair mint you might see the be numbers of Heads on the 10 days. [ Sampling in R. ]

set.seed(2022)
x = rbinom(10, 1000, .5)
x
[1] 488 488 502 493 495 490 524 487 500 506

If person does n’t know you used a fair mint, then they might use the procedure prop.test to see whether the Heads probability each day might be $ 0.5. $

prop.test(x, rep(1000,10), p=rep(.5,10))

        10-sample test for given proportions 
        without continuity correction

data:  x out of rep(1000, 10), null probabilities rep(0.5, 10)
X-squared = 4.988, df = 10, p-value = 0.892
alternative hypothesis: two.sided
null values:
 prop 1  prop 2  prop 3  prop 4  prop 5  prop 6  prop 7  prop 8  prop 9 prop 10 
    0.5     0.5     0.5     0.5     0.5     0.5     0.5     0.5     0.5     0.5 
sample estimates:
 prop 1  prop 2  prop 3  prop 4  prop 5  prop 6  prop 7  prop 8  prop 9 prop 10 
  0.488   0.488   0.502   0.493   0.495   0.490   0.524   0.487   0.500   0.506 

The null hypothesis is that $ p=P ( \mathrm { Heads } ) = 0.5 $ each day, against the option hypothesis that $ p \ne 0.5 $ on one or more of the days. $ H_0 $ can not be rejected because the P-value $ 0.892 > 0.05 = 5\ %. $

intelligibly, not all daily proportions of Heads are precisely $ 0.5, $ but the random variation ( among observed proportions, $ 0.487 $ to $ 0.524 ) $ is not more than would be expected by probability .
By line, if you got results as in the vector y below, then person might wonder whether you used a fair mint each day.

set.seed(228)
y = rbinom(10, 1000, seq(.2, .8, len=10))
y
[1] 182 250 332 408 456 508 611 668 772 809

then prop.test can be used to see if Heads probabilities are all the lapp ( the nonpayment null guess, in case nothing else is specified in the stimulation ) .s prop.test ( y, rep ( 1000,10 ) )

    10-sample test for equality of proportions 
    without continuity correction

data:  y out of rep(1000, 10)
X-squared = 1649.3, df = 9, p-value < 2.2e-16
alternative hypothesis: two.sided
sample estimates:
 prop 1  prop 2  prop 3  prop 4  prop 5  prop 6  prop 7  prop 8  prop 9 prop 10 
  0.182   0.250   0.332   0.408   0.456   0.508   0.611   0.668   0.772   0.809 

For this screen, the P-value is very approximate $ 0 $ thus $ H_0 $ is rejected .
Notes : ( 1 ) Both versions of prop.test are roughly equivalent to chi-squared tests. You could use chi-squared tests if you like, but I find the output signal from prop.test to be more informative for stream purposes .
( 2 ) This Answer is based on one potential interpretation of your motion, which I found to be reasonably undefined. If you have something else in mind, please edit your motion to be more specific .

source : https://ontopwiki.com
Category : Finance