# 0.2 Practice tests (1-4) and final exams  (Page 21/36)

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78 . For a chi-square distribution with five degrees of freedom, the curve is ______________.

## 11.3: test of independence

Use the following information to answer the next four exercises. You are considering conducting a chi-square test of independence for the data in this table, which displays data about cell phone ownership for freshman and seniors at a high school. Your null hypothesis is that cell phone ownership is independent of class standing.

79 . Compute the expected values for the cells.

Cell = Yes Cell = No
Freshman 100 150
Senior 200 50

80 . Compute $\frac{{\left(O-E\right)}^{2}}{z}$ for each cell, where O = observed and E = expected.

81 . What is the chi-square statistic and degrees of freedom for this study?

82 . At the α = 0.5 significance level, what is your decision regarding the null hypothesis?

## 11.4: test of homogeneity

83 . You conduct a chi-square test of homogeneity for data in a five by two table. What is the degrees of freedom for this test?

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

84 . A 2013 poll in the State of California surveyed people about taxing sugar-sweetened beverages. The results are presented in the following table, and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a hypothesis test at the 5% significance level.

Ethnic Group \ Response Type Favor Oppose No Opinion Row Total
White / Non-Hispanic 234 433 43 710
Latino 147 106 19 272
African American 24 41 6 71
Asian American 54 48 16 118
Column Total 459 628 84 1171

85 . In a test of homogeneity, what must be true about the expected value of each cell?

86 . Stated in general terms, what are the null and alternative hypotheses for the chi-square test of independence?

87 . Stated in general terms, what are the null and alternative hypotheses for the chi-square test of homogeneity?

## 11.6: test of a single variance

88 . A lab test claims to have a variance of no more than five. You believe the variance is greater. What are the null and alternative hypothesis to test this?

## 8.1: confidence interval, single population mean, population standard deviation known, normal

1 . $\frac{\sigma }{\sqrt{n}}=\frac{4}{\sqrt{30}}=0.73$

2 . normal

3 . 0.025 or 2.5%; A 95% confidence interval contains 95% of the probability, and excludes five percent, and the five percent excluded is split evenly between the upper and lower tails of the distribution.

4 . z -score = 1.96;

5 . 41 ± 1.43 = (39.57, 42.43); Using the calculator function Zinterval, answer is (40.74, 41.26. Answers differ due to rounding.

6 . The z -value for a 90% confidence interval is 1.645, so EBM = 1.645(0.73) = 1.20085.
The 90% confidence interval is 41 ± 1.20 = (39.80, 42.20).
The calculator function Zinterval answer is (40.78, 41.23). Answers differ due to rounding.

7 . The standard error of measurement is:

The 95% confidence interval is 41 ± 1.12 = (39.88, 42.12).
The calculator function Zinterval answer is (40.84, 41.16). Answers differ due to rounding.

#### Questions & Answers

What is the variances of 568
friend
what variance would have a single value..?
friend
variance happened only in a group of values..
friend
if we have a group of values...1st we find its average..ie..'mean'..then we calculate each value's difeerence from the mean..then we will square each 'difference value'.then we devide total of sqared value by n or n-1..that is what variance...
friend
What is the variances of 258
66,564
Mampy
what is the sample size if the degree of freedom is 25?
26..
friend
25
Tariku
27
Tariku
degrees of freedom may differ with respect to distribution...so tell which distribution you have selected...?
friend
my distribution is 27
Tariku
how to understand statistics
you are working for a bank.The bank manager wants to know the mean waiting time for all customers who visit this bank. she has asked you to estimate this mean by taking a sample . Briefly explain how you will conduct this study. assume the data set on waiting times for 10 customers who visit a bank. Then estimate the population mean. choose your own confidence level.
what marriage for 10 years
fit a least square model of y on x ? what is the regression coefficient ? x : 2 3 6 8 9 10 y : 5 6 7 10 8 11
how can we find the expectation of any function of X?
Jennifer
I've been using this app for some time now. I'm taking a stats class in college in spring and I still have no idea what's going on. I'm also 55 yrs old. Is there another app for people like me?
Tamala
Serious
Hamza
yes I am. it's been decades since I've been in school.
Tamala
who are u
zaheer
is there a private chat we can do
Tamala
hello how can I get PDF of solutions introduction mathematical statistics ( fourth education) who can help me
ahssal
can anyone help me
Halim
what is probability
simply probability means possibility.. definition:Probability is a measure of the likelihood of an event to occur.
laraib
fit a least square model of y on x ? what is the regression coefficient ? x : 2 3 6 8 9 10 y : 5 6 7 10 8 11
Nayab
classification of data by attributes is called
qualitative classification
talal
tell me details about measure of Dispersion
Halim
Following data provided Class Frequency less than 10 10-20 5 15 10-30 25 12 40 and above Which measure of central tendency would you compute and why?
a box contains a few red and a few blue balls.one ball is drawn randomly find the probability of getting a red ball if we know that there are 30 red and 40 blue balls in the box
3/7
RICH
Total=30+40=70 P(red balls) =30/70 Therefore the answer is 3/7
Anuforo
define transport statistical unit
describe each transport statistical unit
Dennis
explain uses of each transport statistical unit
Dennis
identify various transport statistical units with their example
Dennis
I didn't understand about Chi- square.
explain the concept of data analysis and data processing
the following data represent the number of pop up advertisement received by 10 families during the past month.calculate the mean number of advertisement received by each family during month.43,37,35,30,41,23,33,31,16,21
mean=43+37+35+30+41+23+33+31+16/10 =310/10 =31
Anuforo
43+37+35+30+41+23+33+31+16 divided by 10 =310/10 =31
Anuforo
=310/10 =31
Anuforo