Question

The technology committee at a college has stated that the average time spent by students per lab visit has​ increased, and the increase supports the need for increased lab fees. To substantiate this​ claim, the committee randomly samples 12 student lab visits and notes the amount of time spent using the computer. The times are given in the accompanying table.

Time​ (min.)
74	73
67	78
77	146
60	58
65	5
66	57

The previous mean amount of time spent using the lab computer was 63 minutes. Find a​ 95% confidence interval for the true mean. Make a conclusion about the claim. If there are​ outliers, find intervals with and without the outliers present.

The​ 95% confidence interval including any outliers that may exist is ​(____,____) minutes. ​

(Round to one decimal place as​ needed.)

Answer 1

Here, Times are given in the accompanying table. 74 73 67 77 60 65 66 78 146 58 5 S7 & For Finding mean, Sum of all observation Total No. of observation 74 +67+ 77 +60+ 65+66+ 7+ 78 + x = 146+58 +5+57 12 826 68.83 12 Now, We have to find variance, for finding confidence Interval: {(x-x) 2 o - é [(74–68.83)+167-681899+(77-68.83) 12 + (60-68,83)2+(65-68.83) + ( 66-68.83)2+ +(73-68.83)2+(78-68-83)2 +(146-68.83)2 + ( 58-68-83)2+(5-68-83)4+ (57-68.83) *]

10 [ 26.728 + 3.348 + 66.748+ 37.968+ 14.668 + 8.008 + 17.388+ 84.0887 5955. 208 + 117.288+ 4074.268 + 139.9487 I X 10585.656 12 882./38 29.70 ºla NOW, For 95./. confidence Interval we take zx = = 40 96 Zo.05 Then, 12, n = 29.70, 68.83 =) x = 2000s for ㅋ 68.83 ± 1.96 x 29.70 V2 68.83 + 58.212 3.46 68.83 I 16.824 .. upper bound 68.83 + 16.824 85.654 = 85.7 68.83 - 16.8 24 . Lower bound 52.00 = 52.00 Here, The 95%. c. I inche ncluding any outliers that that may exist is 52.0 85.7