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Ho: The distributions of the two populations are the same_ Ha: The distributions of the two populations are not the same 0 = 0.10 The test statistic, Xi = 21.480. C...

Question

Ho: The distributions of the two populations are the same_ Ha: The distributions of the two populations are not the same 0 = 0.10 The test statistic, Xi = 21.480. Critical value: Xi,1o 4.605Which is the correct conclusion of our homogeneity test; at the 10% sigrificance level?Select all that apply:We should reject Ho.We should not reject Ho:Atthe 1O% significance level there IS suffictert evldence t0 corielude that the distributions of the {wo populatlons are not the sume,Atthe !O% slgnificance

Ho: The distributions of the two populations are the same_ Ha: The distributions of the two populations are not the same 0 = 0.10 The test statistic, Xi = 21.480. Critical value: Xi,1o 4.605 Which is the correct conclusion of our homogeneity test; at the 10% sigrificance level? Select all that apply: We should reject Ho. We should not reject Ho: Atthe 1O% significance level there IS suffictert evldence t0 corielude that the distributions of the {wo populatlons are not the sume, Atthe !O% slgnificance level (here Is not sulficlent evldence to conclude that the distributio olthe Ewo papulations are not the same; Hootn Gnnt eb



Answers

We have provided a distribution and the observed frequencies of the values of a variable from a simple random sample of a population. In each case, use the chil-square goodness-of-fit test to decide, at the specified significance level, whether the distribution of the variable differs from the given distribution. Distribution: 0.2,0.1,0.1,0.3,0.3 Observed frequencies: 9,7,1,12,21 Significance level $=0.10$

All right. We want to perform a chi square goodness of fit test. Given the expected distribution, the observed frequencies listed at a significant level of alpha equals 0.1. Before we can start on the test, let's calculate the relative frequencies for the observations that is divide each of the observations by the total sample size. So we obtained from this the proportions 0.29 point 13.0 point 25.28 And we can use our distribution expected frequencies, these frequencies and our sample size to conduct the test. So moving to the test first, we want to see the hypotheses an alpha level. So the hypotheses are H not the variable has a specified distribution, H. A. It does not have the distribution and alpha equals 0.1. As we mentioned. Next step three, we calculate the chi square value high squared remember is the sum of observed by disaffected square over expected in this case 8.416 Next we calculate the critical value for degree freedom and minus one equals four and the alpha level, given we use a chi square table to obtain value 7.779 And since eight is definitely greater than 7.779 we can conclude chi square is in the critical region, and we can reject the null hypothesis.

All right, we want to perform the chi square goodness of fit test for the specified expected probability distribution. The list of observations shown here for each of the three categories at a significant level of alpha equals 30.1 or 1%. Before we move into the test, let's calculate the relative frequencies of the observations. That is each observation number divided by the total sample size. We can use. Our observed proportions are expected proportions and our sample size to calculate the chi square value for our test later. So let's start off into the test with hypotheses and significance level. The hypotheses are not the variable has a specified distribution, H. A. The variable does not have the distribution and alpha equals 10.1 Next we'll calculate the chi squared tests for our sample. Remember Chi Square is equal to the sum of the observations minus expected value squared all over expected or was thinking in terms of total frequency. So this becomes 4.0. Next we calculate the critical value for degrees of freedom and minus one equals two and the specified alpha 20.1 from a chi square table. We obtain critical value 9.210 since four is less than 9.2, we can conclude chi squared is not in the critical region, and we do not or fail to reject H nine, so we don't reject are no hypothesis.

All right. We want to perform a chi square goodness of fit test for the given expected distribution. Observed frequencies at a significant level of alpha equals 0.5 Before proceeding to the test, let's calculate the relative frequencies by which we have the observations. That is. We divide each of the observed frequencies by the sample size to obtain the following proportion next will conduct the test. So in steps one and two we state the hypotheses and significance level. So are null hypothesis. H not is the variable has a specified distribution point to 20.0.4 point 3.1 H. A. Is that it does not have this distribution and our alpha equals 0.5 As stated in step three we calculate our chi squared value. So our chi square value is some of observations expected square over expected. Where we're thinking in terms of frequencies. So this becomes 14.4 next to calculate the critical value. Using a chi square table for our alpha value and a decree freedom and minus one equals three, we obtain 7.815 Since 14.4 is greater than 7.15 we can say that is greater than the critical region, which allows us to reject the null hypothesis H. Non.

We want to perform the chi square goodness of fit test for the given expected distribution. Observed frequencies and alpha equals 0.5 significance. To start off with, let's convert are observed frequencies to relative frequencies for observations. So now that we've done that we can state our hypotheses are significant level. Our hypotheses are H. In our variable has a specified distribution and H. A. It does not have that distribution. The alpha level is stated as 0.5 already. Next in step three, we want to calculate our chi square value. Remember our chi squared is the sum of expected or rather observe minus expected squared over is expected where we're talking in terms of frequencies. So this becomes 30.39 for this problem. So we have to calculate the critical value for degree of freedom and minus one equals three. Using chi square table we obtain 7.815 next We make the conclusion, since high squared is not in the critical region, we do not reject the null hypothesis.


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