Your can retry this question below Sessance companies are...

Your can retry this question below Sessance companies are interested in knowing the population percent of drivers who abways buckle up before riding in a car. They randomly survey 385 drivers and find that 282 claim to always buckle up. Comtruct a 90% confidence interval for the population proportion that claim to always buckle up. <p < Do not round between steps. Round answers to at least 4 decimal places. Submit Question Question 12 0.6/3 pts 7-10 2 Details Score on last try: 0.6 of 3 pts. See Details for more. Get a similar question You can retry this question below Out of a sample of 600 adults aged 18 to 30, 204 still lived with their parents. Based on this, construct a 90% confidence interval for the true population proportion of adults ages 18 to 30 that still live with their parents. Ghe your answers rounded to 4 decimal places. P

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00:01 Hello everyone this question has given that in the year 1983 87 percentage of 19 year old 19 year old people at driving license after 25 years year later that is in the year 2008 75 percentage of 19 year old people have driving license so this data is collected from taking 200 sample space so the question is we have to find the marginal error for the year 1983 and the interval estimate for the year 1983.

00:36 So first of all the marginal error formula is marginal error equal to z into alpha by 2 square root of p cap 1 minus p cap divided by n.

00:49 Where p cap is from the question at the 1983 87 percent so p cap will be 0 .87 and n is 1000 and alpha is 600 and alpha is since at 95 percentage given that 95 percentage confidence we will be having alpha as as alpha as 1 minus 0 .95 which gives you alpha value 0 .05 then we have to find alpha by 2 alpha by 2 equal to 0 .05 divided by 2 gives you 0 .025 then ez alpha by 2 the critical value for the value 0 .25 0 .025 will be plus or minus 1 .96.

01:34 So this is the critical value.

01:39 So, substituting all the values in the formula, marginal error equal to z alpha by 2 into square root of p cap 1 minus p cap divided by n.

01:49 We will be having z alpha by 2 we got 1 .96 into square root of p cap.

01:57 We got 0 .87.

01:59 So 1 minus 0 .87 .87.

02:02 N is thousand to hundred so solving this we will be getting 0 .019 so we are found the marginal error for the year 19803 is 0 .019 then we have to find the interval of estimate so for interval of for interval of estimate first we find a lower limit lower limit will be p cap minus the marginal error p cap is 0 .87 minus the marginal error is 0 .019 which gives you the value 0 .851 then upper limit upper limit will be p -cap plus the marginal error p -cap is 0 .87 plus marginal error is 0 .019 which gives you 0 .89.

02:48 So the interval of estimate is from 0 .851 to 0 .889.

02:56 The next part is we have to find the marginal error at the year 2008 and the interval estimate...

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