Hisyam Hasbi Ahmadinnahar
21/478483/TK/52726 Statistika A
Tugas 3
1. a.) It’s not immediately clear that the data are
normally distributed since n is only 12, but dividing
it into 8 bins shows more or less a normally shaped
distribution. We can also check the assumption
using skewness of the data. It shows a skewness
value of 0.589 which means it’s slightly skewed
more to the right, so not perfectly normal but I can
only assume it’s normal enough, I guess.
b.) we get that the sample mean is 2259.9, and
Figure 1https://www.socscistatistics.com/descriptive/histograms/
the sample standard deviation is 35.6. n is 12. We
get the confidence interval with the formula,
𝑡𝛼
𝑠
𝐶𝐼 = 𝑥 ± × ( )
2
√𝑛
critical value of the t-distribution with n-1 degrees of freedom and a significance level of α/2.
For a 95% confidence interval and n-1 degrees of freedom, α/2 = 0.025 and tα/2 = 2.131.
35.6
𝐶𝐼 = 2259.9 ± 2.201 × (
) = 2259.9 ± 23.6
√12
So the interval would be (2236.3, 2283.5)
c.) it’s almost the same formula but now it’s minus instead of plus minus, and the t critical would
be 1.796. so, 𝐶𝐼 = 2259.9 − 19.2 = 2240.7
This means that we can be 95% confident that the true population mean is greater than or equal
to 2240.7. So, basically the point b is asking us to give a range that they would be 95% in, but the
point c is just asking us to give the minimum value that they would 95% be above from.
2. We can calculate the mean, 𝑥 to be 88.78 and standard deviation (s) to be 15.99. we can get CI
with the formula, CI = x̄ ± t(α/2, n-1) * (s/√n). We know that α = 0.05 and α/2 = 0.025, and we
find t(α/2, n-1) = 2.021.
CI = 88.78 ± 2.021 * (15.99/√41)
CI = 88.78 ± 4.05
So, CI is (84.73, 92.83).
The assumption is that the data is normally distributed.