A gardener plants some tomato seeds, and after 30 days measures the heights (in
ID: 3293284 • Letter: A
Question
A gardener plants some tomato seeds, and after 30 days measures the heights (in cm) of a random sample of 100 of the seedlings. The data are recorded in the file tomatoes.txt.
a) (3 marks) Assuming the seedling heights are iid, provide an (approximate) 98% confidence interval for the mean height.
b) (1 mark) Which result allows you to make the required approximation in a)?
c) (2 marks) Provide a formal interpretation of your confidence interval in a).
as I can not upload a file so the tomatoes.txt file looks like this
height
9
9.8
9.5
10.3
10.1
9.4
12
10.6
9.1
11.8
9.8
11.2
9.3
9.1
9.9
8
11.3
9
11.2
9.8
10.1
11.2
9.3
12.2
9.1
9.9
10.7
10.8
10
10.3
9.5
10.5
9.5
8.6
9.6
11
7.4
7.4
9.8
8.6
10.5
10.2
10.1
10.2
8.4
8.6
10.5
8.5
10.1
8.5
11
10.1
8.7
7.9
10.6
9.9
10.3
9.9
8.5
10.5
11.4
9.2
9.4
8.9
11.4
9.4
9.6
11.7
11.4
9.3
8.9
10
10.3
9.5
9.9
10.3
8.1
10.6
12.2
8.8
10.5
10.5
11.1
9
11.3
9.1
9.9
11.1
8.4
10.9
12.3
10.8
10.3
11.2
11
8.5
10.7
10.5
10.8
11.6
Explanation / Answer
a.
Standard error of the mean = SEM = S/N
N=100, sample size.
alpha=0.02
Confidence interval = m +/- (t(, N-1)*SEM)
calulation
Standard error of the mean = SEM = S/N = 0.109
t(, N-1) = 2.365
Confidence interval = m +/- (t(, N-1)*SEM)
Mean = 9.985
Lower bound: 9.73
Upper bound: 10.24
b. Which result allows you to make the required approximation in a)?
here Normal distribution assumption has not been considered.
so, we use Central LImit theorem which is due to lindberg-levy
c.formal interpretation of your confidence interval
ithe chance of being the original mean outside (9.73,10.24) is only 2%
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