DefinethesetsA1 ={x:<x0}, Ai ={x:i2<xi1}, i = 2,...,7, and A8 = {x : 6 < x < }.

Question

DefinethesetsA1 ={x:<x0}, Ai ={x:i2<xi1}, i = 2,...,7, and A8 = {x : 6 < x < }. A certain hypothesis assigns probabilities pi0 to these sets Ai in accordance with

This hypothesis (concerning the multinomial pdf with k = 8) is to be tested, at the 5% level of significance, by a chi-square test. If the observed frequencies of the sets Ai, i = 1, 2, . . . , 8, are, respectively, 60, 96, 140, 210, 172, 160, 88, and 74, would H0 be accepted at the (approximate) 5% level of significance?

I do not get what is the H0 for this question, and also how to solve the integral of the exp part in this question. Please explain more about these two questions. Thank you.

-(3-3)21 dz, i=1,2, . . . , 7, 8. pi0 = 2 1 2

Explanation / Answer

Here the given integral is cumulative normal distribution function with mean = 3 and = 2

Here for i = 1 ; Pr(- <x0 ) = F(0) = F(<x0 ; 3; 2) = Pr(Z = -3/2 = -1.5) =  0.0668 [ by Z table]

similalry, all p- value are calculated here.

i = 2 ; Pr(i2<xi1} = Pr(0 < x 1) = F(1) - F(0) = Pr(Z < - 1) - Pr(Z < -1.5)

i = 8 ; Pr(6 < x < ) = Pr(6 < x < ; 3; 2) = Pr(Z > 1.5) = 0.0668

Now we will calculate p - value . THe table is given below.

Total values are = sum of all these values = 1000

like that we will calculate the probability for each data set and expected values.

Expected values = 1000 * probbility

Now we will use goodness of fit chisquare statistic here.

H0 :The data given follows the given probability distribution.

Ha : The data given doesn't follow the given probability distribution.  

Now expected and observed table with chi - square statistic

X2 = 6.925

so here degree of freedom = 8 -1 = 7 and alpha = 0.01

so X2critical  = 14.067

so here X2 < X2critical so we shall not reject the null hypothesis and can conclude that the data follows the given distribution.

i x P(x) Expected values (E) 1 0 0.06681 66.81 2 1 0.09185 91.85 3 2 0.14988 149.88 4 3 0.19146 191.46 5 4 0.19146 191.46 6 5 0.14988 149.88 7 6 0.09185 91.85 8 6+ 0.06681 66.81 Sum 1 1000

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