# 0.2 Practice tests (1-4) and final exams  (Page 24/36)

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64 . ${p}_{c}=\frac{{x}_{A}+{x}_{A}}{{n}_{A}+{n}_{A}}=\frac{65+78}{100+100}=0.715$

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 : ${\overline{x}}_{d}\ge 0$
H a : ${\overline{x}}_{d}<0$

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
$\sigma =\sqrt{2\left(df\right)}=\sqrt{40}=6.32$

## 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 $\frac{{\left(O-E\right)}^{2}}{z}$
Enrolled 145 132 145 – 132 = 13 169 $\frac{169}{132}=1.280$
Not enrolled 55 68 55 – 68 = –13 169 $\frac{169}{68}=2.485$

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 $\frac{250\left(300\right)}{500}=150$ $\frac{250\left(200\right)}{500}=100$ 250
Senior $\frac{250\left(300\right)}{500}=150$ $\frac{250\left(200\right)}{500}=100$ 250
Total 300 200 500

80 . $\frac{{\left(100-150\right)}^{2}}{150}=16.67$
$\frac{{\left(150-100\right)}^{2}}{100}=25$
$\frac{{\left(200-100\right)}^{2}}{150}=16.67$
$\frac{{\left(50-100\right)}^{2}}{100}=25$

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

## 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

what is normal distribution
What is the uses of sample in real life
change of origin and scale
3. If the grades of 40000 students in a course at the Hashemite University are distributed according to N(60,400) Then the number of students with grades less than 75 =*
If a constant value is added to every observation of data, then arithmetic mean is obtained by
sum of AM+Constnt
Fazal
data can be defined as numbers in context. suppose you are given the following set of numbers 18,22,22,20,19,21
what are data
what is mode?
what is statistics
Natasha
statistics is a combination of collect data summraize data analyiz data and interprete data
Ali
what is mode
Natasha
what is statistics
It is the science of analysing numerical data in large quantities, especially for the purpose of inferring proportions in a whole from those in a representative sample.
Bernice
history of statistics
statistics was first used by?
Terseer
if a population has a prevalence of Hypertension 5%, what is the probability of 4 people having hypertension from 8 randomly selected individuals?
Carpet land sales persons average 8000 per weekend sales Steve qantas the firm's vice president proposes a compensation plan with new selling incentives Steve hopes that the results of a trial selling period will enable him to conclude that the compensation plan increases the average sales per sales
Supposed we have Standard deviation 1.56, mean 6.36, sample size 25 and Z-score 1.96 at 95% confidence level, what is the confidence interval?
if Y=a+bX and X=c+dY the show that |r|= √hd where r is regression coefficient
this is a linear function. I presume this will be solved simultaneously?
no
Naheed
how can I get esyer statistic?
yes
pakistan
msc completed
i am bba students at nfc
Hamdan
stat. is the subject in bba .... exam is online . .. which fee u charge to slove my exam ?
Hamdan
which uni u completed msc?
Hamdan
no charges. i am just helping you. not for fees
really
Hamdan
yeap
i am so glad this type of people lived in pakistan💔
Hamdan
but unfortunately bba students just live for money🤣
Hamdan
no purpose of life without money🤠
Hamdan
money is not everything
Hamdan
ok ye tu easy topic h bhot
main tume is se related ques aur theory bejti ho
in an examination 60% passed in physics 52% passed in statistics. while 32% failed in both the subject's using relations between class frequencies in attributes find the percentage of student passed in both the subject's
Satish
Hamdan
apply
Hamdan