Question

The annual earnings​ (in dollars) of 35 randomly selected microbiologists are shown in the data table. Use the data to​ (a) find the sample​ mean, (b) find the sample standard​ deviation, and​ (c) construct a​ 98% confidence interval for the population mean.

100,700 81,419 77,543 68,099 51,620 67,741 94,804 66,410 79,448 73,448 44,319 86,757 60,644 58,151 54,634 78,891

47,896 98,388 80,793 92,543 63,410 75,100 50,972 60,936 91,207 84,375 80,002 63,941 74,232 56,985 46,910 89,548

76,023 62,168 82,712

Answer 1

The table given below ,

X
100700	819155344.4
81419	87233545.21
77543	29854093.93
68099	15841275.61
51620	418575182
67741	18819198.37
94804	516420625.5
66410	32138808.19
79448	54300539.83
73448	1873859.832
44319	770623707.2
86757	215440454.9
60644	130761740.7
58151	193992248.2
54634	304331862.9
78891	46401845.37
47896	584822809.3
98388	692157693
80793	75931878.93
92543	418770793.9
63410	75153468.19
75100	9125776.392
50972	445510092.6
60936	124168900.5
91207	365876175.9
84375	151188910.9
80002	62772185.95
63941	66228834.37
74232	4634935.352
56985	227832156.7
46910	633484098.2
89548	305162117.8
76023	15554268.33
62168	98230101.43
82712	113058349.8

From table ,

a) The sample mean is given by ,

b) The sample standard deviation is given by ,

c) The degrees of freedom is given by ,

d.f.=n-1=35-1=34

The critical value is given by ,

; From excel "=TINV(0.02,34)"

Therefore , the 98% confidence interval for the population mean is given by ,