Question

The following is a random sample of n = 90 undergraduate students' annual textbook expense.
460	450	540	770	320	770	680	500	550
260	270	630	560	370	190	500	480	680
580	590	360	380	280	530	580	470	360
600	580	200	360	500	760	610	240	150
600	650	440	360	490	670	340	630	280
220	740	290	260	740	580	520	470	220
360	290	190	230	430	430	310	300	250
660	380	780	800	660	600	380	500	210
370	500	390	310	280	530	510	340	710
640	380	350	800	520	560	560	390	430
2	The variance of the sample is?
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?
5	The lower and upper ends of the 95% confidence interval are:
a

Answer 1

2                      The variance of the sample is?                                                                                        

= 28628.7640

3                      To build an interval estimate of the population mean, the standard error of the mean is?    

                                                                                                                                 

standard error (se) = sqrt( variance /n) = sqrt(28628.7640/90)

=17.8353

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

                        T value with 89 df at 95% level=     1.987                                                                                          

margin of error = t*se =1.987*17.8353 =35.4387

5                      The lower and upper ends of the 95% confidence interval are:                                   

Mean =466.0000

Lower limit = mean-se =466-35.4387 =430.5613

upper limit = mean+se =466+35.4387 =501.4387