Assignment 2, Part A

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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…

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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.

Assignment 2, Part A
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