A coin is thrown until a head occurs and the number X of tosses recorded. After repeating the experiment 256 times, we obtained the following results:x 1 2 3 4 5 6 7 8f 136 60 34 12 9 1 3 1Test the hypothesis, at the 0.05 level of significance, that the observed distribution of X may be fitted by the geometric distribution g(x ; 1 / 2), x=1,2,3

The hypothesis, at the 0.05 significance level,

H₀ : the observed distribution of X may be fitted by the geometric distribution

Hₐ : the observed distribution of X may not be fitted by the geometric distribution

we conclude p-Value > α so, cannot reject null hypothesis. The observed distribution of X may be fitted by the geometric distribution g(x ; 1 / 2), x=1,2,3

A coin is thrown until a head occurs and the number X of tosses recorded.

The probability that head occur on toss ,p = 0.5

Consider the null and alternative hypothesis,

H₀ : the observed distribution of X may be fitted by the geometric distribution

Hₐ : the observed distribution of X may not be fitted by the geometric distribution

The chi square test statistic here as:

χ² = ∑( Oᵢ - Eᵢ)²/Eᵢ

where Oᵢ--> observed value

Eᵢ --> expected value

and the expected frequencies for each X here E(x) = P(X = x) = 256×g(x)

The probability mass function of geometric distribution with parameter 1/2 is g(x) = (1/2)(1 - 1/2)⁽ˣ⁻¹⁾

where x = 0, 1,2 ,3,......

Observed values are f :136 60 34 12 9 1 3 1

Using the data,we calculate the expected values

E₁ = P( x = 1) = 256×g(1) = 256×(1/2)(1/2)⁰

= 128

E₂ = P( x =2) = 256×g(2) = 256×(1/2)(1/2)¹

= 64

E₃ = P( x = 3) = 256×g(3) = 256×(1/2)(1/2)² = 32

E₄ = P( x = 4) = 256×g(4) = 256×(1/2)(1/2)³

= 16

E₅ = P( x = 5) = 256×g(5) = 256×(1/2)(1/2)⁴

= 8

E₆ = P( x = 6) = 256×g(6) = 256×(1/2)(1/2)⁵

= 4

E₇ = P( x = 7) = 256×g(7) = 256×(1/2)(1/2)⁶

= 2

E₈ = P( x = 8) = 256×g(8) = 256×(1/2)(1/2)⁷

= 1

Hence, χ² = ∑( Oᵢ - Eᵢ)²/Eᵢ

= (136 - 128)²/128 + (60 - 64)²/64 + (34 - 32)²/32 + (12-16)²/16 +(9 - 8)²/8 + (1 - 4 )²/4 + (3 - 2)²/2 +(1 - 1)²/1

= 0.5 + 0.25 + 0.125 + 1 + 0.125 + 2.25 + 0.5 =4.75

Significance level is α =0.05

Using the χ² table, the p- value for χ² is 0.6290 .. Now P value > 0.05 , therefore the test is not significant here and we cannot reject the null hypothesis here. Therefore we have sufficient evidence here that the frequencies given are fitted by the geometric (p = 0.5) distribution.

To learn more about Chi-Square test, refer:

https://brainly.com/question/4543358

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Lion (miles per hour)
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A. 19
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