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An observed frequency distribution is 88 tollows Number ol succ83508 Frequancy Let X= tha numbar ot successea In the samples. Use 0.01 significance level t0 test th...

Question

An observed frequency distribution is 88 tollows Number ol succ83508 Frequancy Let X= tha numbar ot successea In the samples. Use 0.01 significance level t0 test the clalm that the obsarved frequencies fit binomial distribution tor which n-3 and p-2/3. To perform the test; Ihe expected value for each category ia requirad Io be at lenst How many catogorios aro satisfied with that condition?Seloct one:

An observed frequency distribution is 88 tollows Number ol succ83508 Frequancy Let X= tha numbar ot successea In the samples. Use 0.01 significance level t0 test the clalm that the obsarved frequencies fit binomial distribution tor which n-3 and p-2/3. To perform the test; Ihe expected value for each category ia requirad Io be at lenst How many catogorios aro satisfied with that condition? Seloct one:



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: 0.5,0.3,0.2 Observed frequencies: 45,39,16 Significance level $=0.01$

Yeah. We want to perform the chi square goodness of fit test for the given expected probability distribution, observed values listed here at a confidence level of alpha equals 0.1 or 1% significance. Before we dive into the test, let's calculate the observation proportions. That is each observation number divided by the sample size. We use these proportions the expected probability distribution and our sample size to calculate R squared value for the test later. So let's dive into the test. The first steps are declaring the hypotheses and confidence level the hypotheses, H not the variable has specified distribution H. A. It does not at alpha equals 0.1 Which we already discussed. Step three we calculate the chi square statistic for this test. Chi square is simply the sum of observed and expected frequency squared, all divided by expected frequencies in this case 9.905 Next Step four We calculate the critical value for degree freedom and minus one equals two in the specified alpha. From a chi square table, we obtain 9.210 Since 9.905 is greater than 9.210 We can conclude that chi squared is in the critical region, which means that we can successfully reject the null hypothesis agent.

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.

Yeah, we want to perform the chi square goodness of fit test for the expected probability distribution listed here, The observed frequencies in a sample of observations at a significant level of alpha equals 0.1. Before moving into the test, let's calculate the observed proportions. That is each observation number divided by the sample size. We use these proportions with are expected distribution and our sample size to calculate the chi square value later. So moving to the test steps one and two are state the hypotheses in alpha we have age, not that the variable has specified distribution and a. J. It does not as well as the confidence level alpha equals 20.1. Next will calculate the chi square value for our test. High square is the some observation minus suspected squared over expected in this case. 5.76 Next we'll turn our critical value for degree of freedom and minus one equals four. And our alpha level specified from a chi square table, we obtain critical value 7.779 Since our chi squared 5.76 is less than this critical value. We can conclude that chi squared is not in the critical region, and we fail to reject the null hypothesis H nine.

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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