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Procedure meets all of tne conditions of a binomial disirbulion except the num ber of Inals not Iixed then tne oeomebc disinculcn can u5e d The probability 0f gelri...

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Procedure meets all of tne conditions of a binomial disirbulion except the num ber of Inals not Iixed then tne oeomebc disinculcn can u5e d The probability 0f gelring the first success Ine xuntrial i5 qiven Plx) = p(T - p)* wnere the probabllty Bucces? any one trial Subjects are randomly selected tor a healtn survey: probabllity Thalsomeone universal Conor (wiln 9rCTC type Rh negau e Nooo) 0.05. Find the probablllty theatthe Ilrst & ublecttobe univurealblond Joncr the Iourlh parson selecledT

procedure meets all of tne conditions of a binomial disirbulion except the num ber of Inals not Iixed then tne oeomebc disinculcn can u5e d The probability 0f gelring the first success Ine xuntrial i5 qiven Plx) = p(T - p)* wnere the probabllty Bucces? any one trial Subjects are randomly selected tor a healtn survey: probabllity Thalsomeone universal Conor (wiln 9rCTC type Rh negau e Nooo) 0.05. Find the probablllty theatthe Ilrst & ublecttobe univurealblond Joncr the Iourlh parson selecled Tne probacuiy (Rqund our ducimapacu& nctdedal



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Determine whether the given procedure results in a binomial distribution (or a distribution that can be treated as binomial. For those that are not binomial, identify at least one requirement that is not satisfied. In an $A A R P$ Bulletin survey, 1019 different adults were randomly selected without replacement. Respondents were asked if they have one or more credit cards, and responses were recorded as "yes" and "no."

In this question. We want to conduct a chi square test of variance and construct confidence intervals as follows. We have a population acts with variance. Sigma squared equals 1 36.2 and around sample. Besides an equal eight and samples identity, aviation or experience 1 15.1 in one. We test the claims sigma squared is less than 1 36.2 at 1% significance. So we have to proceed through a three to solve for this test first. And a we state the alpha level of hypotheses offers the confidence of significant so all vehicles 30.1 H alpha sigma squared equals 1 36.2 H. A single square is less than 1 36 point to our claim. Next. And be we calculate the chi square value degree of freedom and state the assumptions we're making for our distribution this is chi squared equals m minus one squared over signal squared or 5.92 The degree freedom is n minus one equals seven. And we assume x is normally distributed. Next we estimate the p value and see using our chi squared value for the appropriate degree freedom seven from a chi square table. We obtained P between 70.1 point nine. Thus we conclude in D any that we fail to reject asian off because P is greater than alpha, which ultimately means we lack evidence to support the alternative hypothesis AJ Now we can instruct our 90% confidence interval for Sigma Square. So the confidence interval formula is on the left. We have NNS square but we still need excuse square, nextel square, excuse square next elsewhere to the chi square value for our degree of freedom that split The two tails of the chi squared distribution and the values of .05 or areas of .05. This gives a few squared equals 14.7 X squared equals 2.17 from a table. Thus, we conclude that our interval is 57.3 is less than signal square is less than 371.3.

So we're looking for the probability of this binomial experiment with five independent trials with probability success of 0.7 and we're looking for the probability that we have all successes. So we're gonna have 5 to 5, which is going to just be this, which is just gonna be one. And we'll have 0.7 to the fifth and 0.3 to the zeroth power, which is again just one. So we'll do 0.7 to the fifth, which is equal to zero point 16 807

We want to conduct a chi square test variants and construct a confidence interval as follows. We have a population acts with variance, sigma squared equals 47.1. A random sample we obtained from, the population has n equals 15. For sample size and a square example variants equals 83.2. And why we want to test the claim sigma squared is greater than 47.1 at 5% significance. We proceed through steps a through it, it's all first and a week later, Alpha, which is our confidence or significance level in our hypotheses are known variants in our claim. Thus we have alpha equals 0.5 H nine. Sigma squared equals 47.1. H. A single square is greater than 47.1 and b we calculate our chi square value our degree of freedom and state the assumptions we're making a better distribution. This gives chi squared equals m minus one squared over sigma squared equals 24.7304 The degree freedom is m minus one equals 14. And we're assuming X is normally distributed. Next. We estimate the P value based on archive square value, our degree freedom from a chi square table. This gives P approximately 140.25 or less than 0.25 Thus we can conclude that we reject H not because P is less than equal to alpha, which means that we have evidence to support the alternative provinces H N. Next and two. We constructed 95% confidence interval for the variance sigma square. This interval has the formula on the left. We already know and square where we need X U squared. Nextel square X you an Excel squared or the chi squared values that split archive square distribution, distribution of degree freedom 14 and the two tails with value 20.25 for 95% confidence. Thus excuse square in Excel Square to find on the right taken from a chi square table, And we can conclude that 44.6 is less than chi squared is less than 206.9.

Okay, so we have a binomial experiment with five trials, and the success chance is 0.7, and we're looking for the chance of getting exactly one success, which is going to look like this. This is our five choose for which is just gonna end up being five, and we'll have a 0.7. This is for exactly one success. So this is going to be the the one power, then what? 0.3 to the fourth. So if we d'oh 0.3 to the fourth, we end up getting to your 0.0 81 time, 0.7 times five times point your time. 2.7 times five. We get this equal to 0.0 2835 for our finally


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