STATISTC 111 Lecture 9: Lecture 9 Feb 21, 2019

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21 Feb 2019

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4 4 10 13 22,28 31,46 l i l y 4. 5. The relationship between the two variables either deterministic or statistical lists is b has variability b doensthave variability straight line. X determines y y determines x unless horizontalline. I perfect positive correlation b as b deterministic increases y increases perfect negative correlati as x increases y decreases b deterministic o o o no o o strong positive correlation. The linear correlation coefficient r a statistic the corresponding parameter is p this measures between. I e rel the strength of the linear relationship two variables rt use linreglatbx on calc tofind r. Is the linear relationship significant the relationship is significant we can predict y given x. If byfinding the equation of the regression line y ax b. For l1 and l 2 ng o 52. 100 y gxtb these areparameters i y of. Effects o o o o o a o.

Related Questions

Problem 1 (2 points). Every month a clothing store conducts an inventory and calculates the losses due to theft. The store would like to reduce these losses and is considering two methods to do so: the first is to hire a security guard and the second is to install cameras. The manager of the store has conducted the experiment: During the first 6 months he hired a security guard (method 1) and during the next 6 months he installed cameras (method 2). The monthly losses were reported as follows:

Security guard 350 289 407 398 470 259 Cameras 488 301 276 380 421 425

a/ Test with a=0.05 to determine whether there is enough evidence to conclude that expected monthly loss when installing cameras differs from 382. b/Since the cameras are cheaper than the guard, the manager would prefer to install the cameras unless there was enough evidence to infer that the guard was better. Conduct a hypothesis testing to advise what the manager should do? (Assuming equal variances and using a= 0.05)

Problem 2 (2 points). A supermarket chain performed a survey to help determine desirable locations for its new stores. The management of the chain wanted to know whether a linear relationship exists between weekly take home pay (s) and weekly food expenditures (y). A random sample of eight households produced the data shown in the following table Household 1 3 7 8 2 4 5 6 550 450 560 340 510 710 470 660

170 100 160 80 120 190 140 140 Y

a/ Find the least square regression line.

b/Perform the test Ho: Bi=0, H4: B10 (using a = 0.05) to determine whether there is enough evidence to infer that a linear relationship exists between weekly take home pay (s) and weekly food expenditures (y).

champagnelocust794

1.Write a function encipher(s, n) that takes as inputs an arbitrary string s and a non-negative integer n between 0 and 25, and that returns a new string in which the letters in s have been "rotated" by n characters forward in the alphabet, wrapping around as needed. For example:

>>> encipher('hello', 1) result: 'ifmmp'

>>> encipher('hello', 2) result: 'jgnnq'

>>> encipher('hello', 4) result: 'lipps'

Upper-case letters should be "rotated" to upper-case letters, even if you need to wrap around. For example:

>>> encipher('XYZ', 3)

result: 'ABC'

Lower-case letters should be "rotated" to lower-case letters:

>>> encipher('xyz', 3)

result: 'abc'

Non-alphabetic characters should be left unchanged:

>>> encipher('#caesar!', 2)

result: '#ecguct!'

Hints/reminders:

You can use the built-in functions ord and chr convert from single-character strings to integers and back:

>>> ord('a')

result: 97

>>> chr(97)

result: 'a'

You can use the following test to determine if a character is between 'a' and 'z' in the alphabet:

if 'a' <= c <= 'z':

A similar test will work for upper-case letters.

We recommend writing a helper function rot(c, n) that rotates a single character c forward by n spots in the alphabet. We have given you a template for this helper function in ps3pr3.py that checks to ensure that c is a single-character string. We wrote rot13(c) in lecture; rot(c, n) will be very close to rot13(c)! You can test your rot(c, n) as follows:

>>> rot('a', 1)

result: 'b'

>>> rot('y', 2)

result: 'a'

>>> rot('A', 3)

result: 'D'

>>> rot('Y', 3)

result: 'B'

>>> rot('!', 4)

result: '!'

Once you have rot(c, n), you can write a recursive encipher function.

Once you think you have everything working, here are three more examples to try:

>>> encipher('xyza', 1)

result: 'yzab'

>>> encipher('Z A', 2)

result: 'B C'

>>> encipher('Caesar cipher? I prefer Caesar salad.', 25)

result: 'Bzdrzq bhogdq? H oqdedq Bzdrzq rzkzc.'

2.Write a function decipher(s) that takes as input an arbitrary string s that has already been enciphered by having its characters "rotated" by some amount (possibly 0). decipher should return, to the best of its ability, the original English string, which will be some rotation (possibly 0) of the input string s. For example:

>>> decipher('Bzdrzq bhogdq? H oqdedq Bzdrzq rzkzc.')

result: 'Caesar cipher? I prefer Caesar salad.'

Here are two more examples:

>>> decipher('Hu lkbjhapvu pz doha ylthpuz hmaly dl mvynla lclyfaopun dl ohcl slhyulk.')

result: 'An education is what remains after we forget everything we have learned.'

>>> decipher('python')

result: 'eniwdc'

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def rot(c, n):
""" your docstring goes here """
# check to ensure that c is a single character
assert(type(c) == str and len(c) == 1)

def letter_prob(c):
""" if c is the space character (' ') or an alphabetic character,
returns c's monogram probability (for English);
returns 1.0 for any other character.
adapted from:

"""
# check to ensure that c is a single character
assert(type(c) == str and len(c) == 1)

if c == ' ': return 0.1904
if c == 'e' or c == 'E': return 0.1017
if c == 't' or c == 'T': return 0.0737
if c == 'a' or c == 'A': return 0.0661
if c == 'o' or c == 'O': return 0.0610
if c == 'i' or c == 'I': return 0.0562
if c == 'n' or c == 'N': return 0.0557
if c == 'h' or c == 'H': return 0.0542
if c == 's' or c == 'S': return 0.0508
if c == 'r' or c == 'R': return 0.0458
if c == 'd' or c == 'D': return 0.0369
if c == 'l' or c == 'L': return 0.0325
if c == 'u' or c == 'U': return 0.0228
if c == 'm' or c == 'M': return 0.0205
if c == 'c' or c == 'C': return 0.0192
if c == 'w' or c == 'W': return 0.0190
if c == 'f' or c == 'F': return 0.0175
if c == 'y' or c == 'Y': return 0.0165
if c == 'g' or c == 'G': return 0.0161
if c == 'p' or c == 'P': return 0.0131
if c == 'b' or c == 'B': return 0.0115
if c == 'v' or c == 'V': return 0.0088
if c == 'k' or c == 'K': return 0.0066
if c == 'x' or c == 'X': return 0.0014
if c == 'j' or c == 'J': return 0.0008
if c == 'q' or c == 'Q': return 0.0008
if c == 'z' or c == 'Z': return 0.0005
return 1.0