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Vasya Sidorov claims that he goes to the cinema twice as often as to the gym, and to the gym twice as often as to the theater. Over the past six months, he has been to the theater 10 times, to the gym 17 times and to the cinema 39 times. At the level significance of 5% check Vasya's statement.

Seems that theoretical means would be 20 and 40. Seems that statement is true, if both substatement are true:
A) Goes to gym twice often than theatre
B) Goes to cinema 4 times often than in theater.
$n = 66$, then
answer Hypothesis test of two proportions where one proportion is k times the other one

I got that A and B are true, but I'm not doubt that my solution is right. I would be grateful for your help!

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There are several similar methods. One is implemented in R as prop.test, a test with the null hypothesis $H_0$ that the probabilities of the three activities are $1/7, 2/7, 4/7,$ respectively. The alternative is that these probabilities are not true.

The data your state are consistent with that null hypothesis. (That is, $H_0$ is not rejected.)

prop.test(c(10,17,39), c(66,66,66),  p = c(1,2,4)/7)

        3-sample test for given proportions
        without continuity correction

data:  c(10, 17, 39) out of c(66, 66, 66), null 
probabilities c(1, 2, 4)/7
X-squared = 0.39874, df = 3, p-value = 0.9405
alternative hypothesis: two.sided
null values:
   prop 1    prop 2    prop 3 
0.1428571 0.2857143 0.5714286 
sample estimates:
   prop 1    prop 2    prop 3 
0.1515152 0.2575758 0.5909091 
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  • $\begingroup$ Thanks you! However do you know how to solve this task by hand (without machine)? $\endgroup$ Apr 28 at 19:02
  • $\begingroup$ Can I use youtube.com/watch?v=pCbNUnZ98oE and my idea, that statement is true, if A and B are true? $\endgroup$ Apr 28 at 19:07
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    $\begingroup$ Slightly different, perhaps more easily computed, method is to chi-sq goodness of fit test with observed frequencies $X_i. i = 1,2,3$ as $10, 17, 39$ for the three activities out of $66$. then find expected frequencies as $E_i = 66(1/7,2/7,4/7).$ Finally use chi-squared test statistic. $H = \sum_i \frac{(X_i-E_i)^2}{E_i} \stackrel{aprx}{\sim}\mathsf{Chisq}(\nu=2).$ $\endgroup$
    – BruceET
    Apr 28 at 19:12
  • $\begingroup$ I do not ordinarily comment on the accuracy/validity of specific You Tube videos. Some are fine, some are not; you have to do your own quality management. $\endgroup$
    – BruceET
    Apr 28 at 19:24
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    $\begingroup$ Thanks you for chisq! $\endgroup$ Apr 28 at 19:40

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