A1. Generate an array x of 10000 random values from
a uniform distribution on the interval [0,20],
then use a for loop to determine the percentage
of values that are in the interval [5,12].
Note: A1 asks for a percentage, not a count and not
a proportion.
A2. Repeat A1 500 times, then compute the average
of the 500 percentages found.
A3. For the array x in A1, use a while loop to determine
the number of random entries required to find the
first that is less than 4.
A4. Repeat A3 1000 times, then compute the average for the
number of random entries required.
A5. For the array x in A1, use a while loop to determine
the number of random entries required to find the
third entry that exceeds 12.
A6. Repeat A5 1000 times, then compute the average for the
number of random entries required.
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Assignment 2, Part B
For this problem you will draw samples from a normal
population with mean 40 and standard deviation 12.
Run the code below to generate your population, which
will consist of 500,000 elements.
import numpy as np
p1 = np.random.normal(40,12,size=500000)
a) The formula for a 95% confidence interval for the
population mean is given by
[xbar - 1.96*sigma/sqrt(n), xbar + 1.96*sigma/sqrt(n)]
where xbar is the sample mean, sigma is the population
standard deviation, and n is the sample size.
i) Select 10,000 random samples of size 10 from p1. For
each sample, find the corresponding confidence
interval, and then determine the proportion of
confidence intervals that contain the population mean.
ii) Repeat part i) using samples of size 20.
iii) Repeat part i) using samples of size 30.
b) Frequently in applications the population standard
deviation is not known. In such cases, the sample
standard deviation is used instead. Repeat part a)
replacing the population standard deviation with the
standard deviation from each sample, so that the
formula is
[xbar - 1.96*stdev/sqrt(n), xbar + 1.96*stdev/sqrt(n)]
Tip: The command for the standard deviation is
np.std(data, ddof=1)
c) Your answers in part b) should be a bit off. The
problem is that a t-distribution is appropriate when
using the sample standard deviation. Repeat part b),
this time using t* in place of 1.96 in the formula,
where: t* = 2.262 for n = 10, t* = 2.093 for n = 20,
and t* = 2.045 for n = 30.
