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7. (7Pts) Assume that recent wedding costs in the United States have a normal di

ID: 2922989 • Letter: 7

Question

7. (7Pts) Assume that recent wedding costs in the United States have a normal distibutoh l a standard deviation of $810 Based on a random sample of 29 recent US weddings, a 95% confidence interval for the mean cost, of all recent US weddings is computed and found to be $22,704.5 to $29,949.3. List each a) Determine if the assumptions were met to compute this confidence interval. assumption met and show the information given or not given to determine if each assumption is b) If the assumptions are met, write the interpretation.

Explanation / Answer

Solution:-) A) The assumptions for using the confidence interval are:-

1) Randomization Condition:- The data must be sampled randomly i.e. one of the good sampling methodologies.

2) Independence Assumption:- The sample values must be independent of each other i.e. the occurence of the event does not influence the event of other.One must be noted that if we choose the random samples means they are independent.

3) Sample Size must be smaller than Population:- When the sample is drawn without replacement (usually the case), the sample size n, should be smaller than 10% of the population.

4) Normality Assumption:- There should be normality assumption that is either the population is normal or you can use the central limit theorem for sample size is large.

The above assumptions should be met to use the confidence interval criteria.As it is given that the sample is drawn from the normal populaton. So it holds the normality assumption. The sample size is small i.e. n=29, hence the sample size is so smaller than the population(recent U.S.Weddings). The samples are drawn randomly i.e. randomization condition also holds and as they are randomly drawn. This implies that they are independent also. Hence, the above sample holds each and every assumption to compute this confidence interval.

2) The average cost of all US recent weddings fall between ($22704.5,$29949.3) with 95% Confidence, means there is chances of 5% error .

CHEERS!!

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