# 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 The Confidence Interval
sample mean 25, sample standard deviation 20, sample size 200, calculate the confidence interval using the given values and the original confidence level of 90%.
Can you help me in mathematical statistics problems?
yes
Kc
Pls who can help me to teach me statistics
nasir
i need tutor for statistics plz
Rana
ok
Ekene
the power of the test is
please can anyone help me solve these questions below? I need help please.
MMSI
a)An investor wants to eliminate seven of the investments in her portfolio by selling 4 stocks and 3 bonds. In how many can these be sold if among 25 securities in the portfolio,13 are stocks and the rest bonds?
MMSI
a)If a random variable has the standard normal distribution,what are the probabilities that it will take on a value: i)Less than 1.64 ii)Greater than-0.47
MMSI
b)A random variable has a normal distribution with a mean of 60 and standard deviation 5.2.What are the probabilities that the random variable will take on a value: i)Less than 65.2 ii)Between 48 and 72?
MMSI
b)If the probability that an individual suffers a bad reaction from injection of a given serum is 0.001,use the Poisson law to calculate the probability that out of 2000 individuals i)Exactly 3 individuals will suffer a bad reaction. ii)More than 2 individuals will suffer a bad reaction.
MMSI
b)The breakfast menu serve data popular 5-star Hotel in Accra consists of the following items: Juice-Mango,Grape,Apple. Toast-Whitewheat,Whole wheat. Egg:Fried,Hard-boiled,Scrambled. Beverage:Coffee,Tea,Cocoa.
MMSI
Continuation of the last question.Assist the Hotel manager to determine the number of possible breakfast combinations that can be served, one from each category
MMSI
MMSI
3x2x3
Vince
Are you answering the last question?
MMSI
MMSI
bias came in sampling due to
sampling error
Vikram
what is the difference between population and sample
Inam
Sample is the group of individual who participate in your study. Sample is a subset of population. Population is the broader group of people to whom you intend to generalize the results of your study.
Ekene
how do you find z if you only know the area of .0808
construct a frequency distribution
Sana
How to take a random sample of 30 observations
you can use the random function to generate 30 numbers or observation
smita
How we can calculate chi-square if observed x٫y٫z/frequency 40,30,20 Total/90
calculate chi-square if observed x,y,z frequency 40,30,20total 90
Insha
find t value,if boysN1, ،32,M1,87.43 S1square,39.40.GirlsN2,34,M2,82.58S2square,40.80 Determine whether the results are significant or insignificant
Insha
The heights of a random sample of 100 entering HRM Freshman of a certain college is 157 cm with a standard deviation of 8cm. test the data against the claim that the overall height of all entering HRM students is 160 cm. previous studies showed that
complete the question.. as data given N = 100,mean= 157 cm, std dev = 8 cm..
smita
Z=x-mu/ std dev
smita
the power of the test is
Ejaz
find the mean of 25,26,23,25,45,45,58,58,50,25
add all n divide by 10 i.e 38
smita
38
hhaa
amit
1 . The “average increase” for all NASDAQ stocks is the:
STATISTICS IN PRACTICE: This is a group assignment that seeks to reveal students understanding of statistics in general and it’s practical usefulness. The following are the guidelines; 1.      Each group has to identify a natural process or activity and gather data about/from the process. 2.
The diameter of an electric cable,say, X is assumed to be continoues random variable with p.d.f f(x)=6x(1-x); ≤x≤1 a)check that f(X) is p.d.f b) determine a number b such that p(Xb)
A manufacturer estimate 3% of his output is defective. Find the probability that in a sample of 10 items (a) less than two will be defective (b) more than two will be defective.
A manufacturer estimates that 3% of his output of a small item is defective. Find the probabilities that in a sample of 10 items (a) less than two and (b) more than two items will be defective.
ISAIAH
use binomial distribution with parameter n=10, p= 0.03, q=0.97
the standard deviation of a symmetrical distribution is 7.8 . what must be the value of forth moment about the mean in order that distribution be a) leptokurtic b) mesokurtic c) platy kyrtic intrept the obtain value of a b and c By By By Stephen Voron By Richley Crapo By Brooke Delaney By OpenStax By Marion Cabalfin By Stephen Voron By Marion Cabalfin By OpenStax By Dan Ariely By IES Portal