Suppose a survey of 500 people age 18 to 34 indicated that 32.2% of them live with one or both of their parents. Calculate and interpret a confidence interval estimate for the true proportion of all people age 18 to 34 who live with one or both parents. Use a 94% confidence level. CHAPTER 8: FLOW CHART VIEW OF FORMULAS FOR CONFIDENCE INTERVAL

The 94% confidence interval would be given by (0.283;0.361)

Step-by-step explanation:

Notation and definitions

random sample taken

estimated proportion of people between 18 to 34 who live with their parents

true population proportion of people between 18 to 34 who live with their parents  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 94% of confidence, our significance level would be given by and . And the critical value would be given by:

The confidence interval for the mean is given by the following formula:  

If we replace the values obtained we got:

The 94% confidence interval would be given by (0.283;0.361)

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

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step-by-step explanation: