# 11.3 Test of independence  (Page 3/20)

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De Anza College is interested in the relationship between anxiety level and the need to succeed in school. A random sample of 400 students took a test that measured anxiety level and need to succeed in school. [link] shows the results. De Anza College wants to know if anxiety level and need to succeed in school are independent events.

Need to succeed in school vs. anxiety level
Need to Succeed in School High
Anxiety
Med-high
Anxiety
Medium
Anxiety
Med-low
Anxiety
Low
Anxiety
Row Total
High Need 35 42 53 15 10 155
Medium Need 18 48 63 33 31 193
Low Need 4 5 11 15 17 52
Column Total 57 95 127 63 58 400

a. How many high anxiety level students are expected to have a high need to succeed in school?

a. The column total for a high anxiety level is 57. The row total for high need to succeed in school is 155. The sample size or total surveyed is 400.

$E=\frac{\text{(row total)(column total)}}{\text{total surveyed}}=\frac{155\cdot 57}{400}=22.09$

The expected number of students who have a high anxiety level and a high need to succeed in school is about 22.

b. If the two variables are independent, how many students do you expect to have a low need to succeed in school and a med-low level of anxiety?

b. The column total for a med-low anxiety level is 63. The row total for a low need to succeed in school is 52. The sample size or total surveyed is 400.

c. $E=\frac{\text{(row total)(column total)}}{\text{total surveyed}}$ = ________

c. $E=\frac{\text{(row total)(column total)}}{\text{total surveyed}}=8.19$

d. The expected number of students who have a med-low anxiety level and a low need to succeed in school is about ________.

d. 8

## Try it

Refer back to the information in [link] . How many service providing jobs are there expected to be in 2020? How many nonagriculture wage and salary jobs are there expected to be in 2020?

12,727, 14,965

## References

DiCamilo, Mark, Mervin Field, “Most Californians See a Direct Linkage between Obesity and Sugary Sodas. Two in Three Voters Support Taxing Sugar-Sweetened Beverages If Proceeds are Tied to Improving School Nutrition and Physical Activity Programs.” The Field Poll, released Feb. 14, 2013. Available online at http://field.com/fieldpollonline/subscribers/Rls2436.pdf (accessed May 24, 2013).

Harris Interactive, “Favorite Flavor of Ice Cream.” Available online at http://www.statisticbrain.com/favorite-flavor-of-ice-cream (accessed May 24, 2013)

“Youngest Online Entrepreneurs List.” Available online at http://www.statisticbrain.com/youngest-online-entrepreneur-list (accessed May 24, 2013).

## Chapter review

To assess whether two factors are independent or not, you can apply the test of independence that uses the chi-square distribution. The null hypothesis for this test states that the two factors are independent. The test compares observed values to expected values. The test is right-tailed. Each observation or cell category must have an expected value of at least 5.

## Test of independence

• The number of degrees of freedom is equal to (number of columns - 1)(number of rows - 1).
• The test statistic is $\underset{\left(i\cdot j\right)}{\Sigma }\frac{{\left(O–E\right)}^{2}}{E}$ where O = observed values, E = expected values, i = the number of rows in the table, and j = the number of columns in the table.
• If the null hypothesis is true, the expected number $E=\frac{\text{(row total)(column total)}}{\text{total surveyed}}$ .

binomial distribution tends to normal distribution
if sample size n very large and probability tends to 0.5 if these both conditions are satisfied by binomial distribution it would tends to normal distribution
sravani
n tends to infinite i.e large Probability tends to 0 i.e indefinitely small. np = lamda
Anji
the above is poison to Bin
Anji
no of trails n tends to indefinitely large..i.e infine neither p nor q is very small Then bin tends to normal
Anji
if the death of of the snow is my yard is normally distributed with the m is equals to 2.5 and what is the probability that a randomly chosen location with have a no that between 2.25 and 2.76
hey
Shubham
🤔
Iqra
hello
Sakshi
hii
Rushikesh
helow
why Statistics so hard
Mohd
ho geya solve
Sakshi
it's not hard
Sakshi
it is hard 😭
Mohd
solution?
Abdul
hii
it's just need to be concentrate
Akinyemi
exactly..... concentration is very important
Iqra
rewrite the question
what is the true statement about random variable?
A consumer advocate agency wants to estimate the mean repair cost of a washing machine. the agency randomly selects 40 repair cost and find the mean to be $100.00.The standards deviation is$17.50. Construct a 90% confidence interval for the mean.
pls I need understand this statistics very will is giving me problem
Sixty-four third year high school students were given a standardized reading comprehension test. The mean and standard deviation obtained were 52.27 and 8.24, respectively. Is the mean significantly different from the population mean of 50? Use the 5% level of significance.
No
Ariel
how do I find the modal class
look for the highest occuring number in the class
Kusi
the probability of an event occuring is defined as?
The probability of an even occurring is expected event÷ event being cancelled or event occurring / event not occurring
Gokuna
what is simple bar chat
Simple Bar Chart is a Diagram which shows the data values in form of horizontal bars. It shows categories along y-axis and values along x-axis. The x-axis displays above the bars and y-axis displays on left of the bars with the bars extending to the right side according to their values.
statistics is percentage only
the first word is chance for that we use percentages
it is not at all that statistics is a percentage only
Shambhavi
I need more examples
how to calculate sample needed
mole of sample/mole ratio or Va Vb
Gokuna
how to I solve for arithmetic mean
Yeah. for you to say.
James
yes
niharu
how do I solve for arithmetic mean
niharu
add all the data and divide by the number of data sets. For example, if test scores were 70, 60, 70, 80 the total is 280 and the total data sets referred to as N is 4. Therfore the mean or arthritmatic average is 70. I hope this helps.
Jim
*Tan A - Tan B = sin(A-B)/CosA CosB ... *2sinQ/Cos 3Q = tan 3Q - tan Q