Question

The following is a random sample of n = 90 undergraduate students' annual textbook expense.
610	600	300	420	520	470	430	520	400
370	730	480	450	500	650	370	540	330
690	550	450	450	750	750	660	700	300
770	760	390	680	450	590	630	530	700
580	390	330	320	350	490	310	320	780
590	370	470	760	550	630	450	640	620
520	440	720	660	440	770	380	450	800
720	370	550	460	560	590	350	790	700
550	650	750	400	490	690	610	730	560
420	410	720	650	450	540	590	310	370

1	The point estimate of the population mean is $540

2	The variance of the sample is 20,809
3	To build an interval estimate of the population mean, the standard error of the mean is _____?

4	The margin of error for a 95% interval estimate is ___?

For one and two, the bold numbers are the answers I got, they may not be correct. However, 3 & 4 is what I really need help with, as I have no clue how to solve this. If you could explain your steps, that would help so much so that I can do it on my own in the future!

Answer 1

0 The population mean: point estimate of the = 540] © The variance of the sample: s” = 20808.9886 ~208091 ☺ n= 90, 5=520808.9886 standard error of the mean : 520808.9886 se = Er = 14 Too =115.2056237 se = © degrees of freedom = n-1 = 90-1= 89 significance level, d=1-0.95 = 0.05 Critical value, tdiz = t0.025 = 1.987 :. margin of error for a 95% interval estimate: moe = td12 * se = 1.987 * 15.205623 =130.21357

** Your calculations for mean and standard deviation are correct. For standard error and margin of error, round the answers as per your requirement. Thank you..