I understand that these will be considered seperate questions, but could I get a full answer please.
Thank you.
1 A civil engineering agency monitors water quality by measuring the amount of suspended solids in a sample of river water. The following results, measured in parts per million, were observed over 60 days:
21.2 |
19.9 |
20.2 |
15.6 |
17.6 |
14.0 |
19.7 |
18.3 |
26.2 |
23.6 |
20.2 |
19.8 |
22.2 |
27.3 |
23.4 |
19.8 |
22.3 |
16.9 |
16.4 |
22.0 |
19.8 |
21.1 |
20.4 |
18.6 |
21.8 |
22.5 |
20.0 |
21.6 |
26.3 |
22.3 |
25.9 |
21.0 |
19.0 |
23.3 |
26.1 |
22.3 |
23.6 |
20.1 |
22.8 |
17.8 |
12.9 |
25.7 |
26.2 |
25.1 |
22.5 |
22.3 |
21.8 |
22.2 |
22.6 |
17.3 |
22.6 |
20.5 |
20.6 |
24.7 |
22.3 |
21.4 |
25.4 |
17.7 |
18.9 |
21.7 |
(a) Plot a histogram of this data, with classes (12.25, 14.25], (14.25, 16.25], …, either manually or using Excel.
(b) Comment on the shape of the distribution of the amount of suspended solids. What measures of location and spread are appropriate for this data set and why? Calculate them.
(c) Assuming that the data can be sufficiently well modelled by a normal distribution, construct a 95% confidence interval for the mean number of suspended solids.
(d) Is there any evidence, at the 95% confidence level, that the mean number of suspended solids exceeds 25 parts per million?
Pleasure of location for given data is mean and spread for data is variance. Because the values are discrete in nature, so, that’s why we use mean and variance here. Total sum of observation is 1277.3. Mean = 21.2883. V(x) = 9.349184 (Calculated using Excel). 95% C.I for mean number of suspended coil. Therefore, 95% C.I for mean number of suspended coil is (20.5147, 22.06198). (d) No, there is no any evidence that at 95% confidence level, that the mean number of suspended solids exceeds 25 parts per million.
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