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64 . p c = x A + x A n A + n A = 65 + 78 100 + 100 = 0.715

65 . Using the calculator function 2-PropZTest, the p-value = 0.0417. Reject the null hypothesis. At the 3% significance level, here is sufficient evidence to conclude that there is a difference between the proportions of households in the two communities that have cable service.

66 . Using the calculator function 2-PropZTest, the p -value = 0.0417. Do not reject the null hypothesis. At the 1% significance level, there is insufficient evidence to conclude that there is a difference between the proportions of households in the two communities that have cable service.

10.4: matched or paired samples

67 . H 0 : x ¯ d 0
H a : x ¯ d < 0

68 . t = – 4.5644

69 . df = 30 – 1 = 29.

70 . Using the calculator function TTEST, the p -value = 0.00004 so reject the null hypothesis. At the 5% level, there is sufficient evidence to conclude that the participants lost weight, on average.

71 . A positive t -statistic would mean that participants, on average, gained weight over the six months.

11.1: facts about the chi-square distribution

72 . μ = df = 20
σ = 2 ( d f ) = 40 = 6.32

11.2: goodness-of-fit test

73 . Enrolled = 200(0.66) = 132. Not enrolled = 200(0.34) = 68

74 .

Observed (O) Expected (E) O – E (O – E)2 ( O E ) 2 z
Enrolled 145 132 145 – 132 = 13 169 169 132 = 1.280
Not enrolled 55 68 55 – 68 = –13 169 169 68 = 2.485

75 . df = n – 1 = 2 – 1 = 1.

76 . Using the calculator function Chi-square GOF – Test (in STAT TESTS), the test statistic is 3.7656 and the p-value is 0.0523. Do not reject the null hypothesis. At the 5% significance level, there is insufficient evidence to conclude that high school most recent graduating class distribution of enrolled and not enrolled does not fit that of the national distribution.

77 . approximates the normal

78 . skewed right

11.3: test of independence

79 .

Cell = Yes Cell = No Total
Freshman 250 ( 300 ) 500 = 150 250 ( 200 ) 500 = 100 250
Senior 250 ( 300 ) 500 = 150 250 ( 200 ) 500 = 100 250
Total 300 200 500

80 . ( 100 150 ) 2 150 = 16.67
( 150 100 ) 2 100 = 25
( 200 100 ) 2 150 = 16.67
( 50 100 ) 2 100 = 25

81 . Chi-square = 16.67 + 25 + 16.67 + 25 = 83.34.
df = ( r – 1)( c – 1) = 1

82 . p -value = P (Chi-square, 83.34) = 0
Reject the null hypothesis.
You could also use the calculator function STAT TESTS Chi-Square – Test.

11.4: test of homogeneity

83 . The table has five rows and two columns. df = ( r – 1)( c – 1) = (4)(1) = 4.

11.5: comparison summary of the chi-square tests: goodness-of-fit, independence and homogeneity

84 . Using the calculator function (STAT TESTS) Chi-square Test, the p -value = 0. Reject the null hypothesis. At the 5% significance level, there is sufficient evidence to conclude that the poll responses independent of the participants’ ethnic group.

85 . The expected value of each cell must be at least five.

86 . H 0 : The variables are independent.
H a : The variables are not independent.

87 . H 0 : The populations have the same distribution.
H a : The populations do not have the same distribution.

11.6: test of a single variance

88 . H 0 : σ 2 ≤ 5
H a : σ 2 >5

Practice test 4

12.1 linear equations

1 . Which of the following equations is/are linear?

  1. y = –3 x
  2. y = 0.2 + 0.74 x
  3. y = –9.4 – 2 x
  4. A and B
  5. A, B, and C

Questions & Answers

How do you get log of normal population
Shan Reply
plz give me answer of this question show that mean of population or sampling distribution are equally
Mari
what is the probability of getting no head face up in three tosses of a fair coin?
Epara Reply
In how many ways can probability be assigned to an event of interest?
Epara
Hey guys can someone help me with combinations and permutations
Lion
Phone lines on an airline system are occupied 50% of time assume that 10 calls are placed to the airline. What is the probability at least 1 call the lines are occupied?
Highness Reply
Phone lines on an airline system are occupied 50% of time assume that 10 calls are placed to the airline. What is the probability at least 1 call the lines are occupied?
Highness Reply
Why is the method of selecting the sample even more important than the size of the sample?
Nana Reply
formulaas for gruoped and ungrouped of quartiles
chatered Reply
Why is the method of selecting the sample even more important than the size of the sample?
Nana
una empresa productora que participa en el mercado de lapices tiene la siguente funcion de demanda Qd=a la raiz de 9 -9p si la elasticidad precio de demanda de la empresa es epd=0,5 determinar a) hallar el precio y cantidad
Jany Reply
find the mean mew and the standard devation sigma of the given population 9.8,10.2,10.4,9.8,10.0,10.2,9.6
Vaneza Reply
1. A card is drawn at random from an ordinary deck of 52 playing cards.* *Find the probability that it is* (a). an ace (b). a jack of hearts (c). a three of clubs or a six of diamonds (d). a heart (e). any suit except hearts (f). a ten or a spade (g). neither a four nor a club. *Hint:* 1st determine how many of the following is in a deck of cards. A deck of cards have 52 cards ♠️- Club ♣️ - Spade ♥️ - Heart ♦️- Diamond
Agness Reply
a measure of central tendency is a typical value around which other figures congregate
Anand Reply
hmm mean mode median
Umar
nothing
farri
Y = alpha0 + alpha1X1 + E what is this equation
Musawenkosi Reply
An estimated linear regression equation
Carlos
Thank you
Musawenkosi
simple linear regression .. where Alpho zero is reg constant ( intercept of the reg line) and Alpha1 is the regression coefficient ( slope of the regression line)
Umar
null and altarnate hypothesis are the statement about
farri Reply
about any population of interest
Umar
don't know
farri
I said we give hypothesis about any population and mean , in null hyp we say sample mean is equal to the population mean where as in alternate hyp we say sample mean is not equal to the pop mean .. to test these things we use students test statistics commonly
Umar
ok thnx
farri
welcome
Umar
probability of getting one black card
kaynat Reply
from a standard deck?
Umar
it will be 1/2
Umar
coz there are 26 blacks card out of 52 in a deck
Umar
so prob of getting a black card out of the deck = 26/52 = 1/5 =0.5
Umar
when are not two events mutually exclusive? 1) they overlap in a venn diagram 2) Probability of one affects the other
Junaid Reply
both
Umar
A stenographer claims that she can take dictation at the rate of 120 words per minute can we reject her claim on the basis of 100 trails in which she demonstrates a mean of 116 words with a variance of 225 words
Aromal Reply
the hypothesis to be tested is the claim to be tested
smita
H0= 120,Ha not equal to 120 x bar =116,s=225,n=100
smita

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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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