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Test the hypothesis two ways (1) using the Chi-square test and (2) using the z-test for independence with significance level of 10%. Show how the two test statistics are related and compare the p-values. This site contains user submitted content, comments and opinions and is for informational purposes only. Apple may provide or recommend responses as a possible solution based on the information provided; every potential issue may involve several factors not detailed in the conversations captured in an electronic forum and Apple can therefore provide no guarantee as to the. Apple Magic Keyboard, Magic Mouse 2, Magic Trackpad 2. Redesigned to be fully rechargeable and even more of a joy to use.

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Perfect Squares and their Square Roots


Perfect Square:

Taking a positive integer and squaring it (multiplying it by itself) equals a perfect square.
Example: 3 x 3 = 9 Thus: 9 is a perfect square.
Taking the square root (principal square root) of that perfect square equals the original positive integer.
Example: 9 = 3 Where: 3 is the original integer.
Note: An integer has no fractional or decimal part, and thus a perfect square (which is also an integer) has no fractional or decimal part.


( Perfect Squares List from 1 to 10,000. )


Positive
Integer
Integer
Squared=
Perfect Squares
List
Square Root of
Perfect Square=
Original
Integer
11 ^2 =1 1 =1
22 ^2 =4 4 =2
33 ^2 =9 9 =3
44 ^2 =16 16 =4
55 ^2 =25 25 =5
66 ^2 =36 36 =6
77 ^2 =49 49 =7
88 ^2 =64 64 =8
99 ^2 =81 81 =9
1010 ^2 =100 100 =10
1111 ^2 =121 121 =11
1212 ^2 =144 144 =12
1313 ^2 =169 169 =13
1414 ^2 =196 196 =14
1515 ^2 =225 225 =15
1616 ^2 =256 256 =16
1717 ^2 =289 289 =17
1818 ^2 =324 324 =18
1919 ^2 =361 361 =19
2020 ^2 =400 400 =20
2121 ^2 =441 441 =21
2222 ^2 =484 484 =22
2323 ^2 =529 529 =23
2424 ^2 =576 576 =24
2525 ^2 =625 625 =25
2626 ^2 =676 676 =26
2727 ^2 =729 729 =27
2828 ^2 =784 784 =28
2929 ^2 =841 841 =29
3030 ^2 =900 900 =30
3131 ^2 =961 961 =31
3232 ^2 =1024 1024 =32
3333 ^2 =1089 1089 =33
3434 ^2 =1156 1156 =34
3535 ^2 =1225 1225 =35
3636 ^2 =1296 1296 =36
3737 ^2 =1369 1369 =37
3838 ^2 =1444 1444 =38
3939 ^2 =1521 1521 =39
4040 ^2 =1600 1600 =40
4141 ^2 =1681 1681 =41
4242 ^2 =1764 1764 =42
4343 ^2 =1849 1849 =43
4444 ^2 =1936 1936 =44
4545 ^2 =2025 2025 =45
4646 ^2 =2116 2116 =46
4747 ^2 =2209 2209 =47
4848 ^2 =2304 2304 =48
4949 ^2 =2401 2401 =49
5050 ^2 =2500 2500 =50
5151 ^2 =2601 2601 =51
5252 ^2 =2704 2704 =52
5353 ^2 =2809 2809 =53
5454 ^2 =2916 2916 =54
5555 ^2 =3025 3025 =55
5656 ^2 =3136 3136 =56
5757 ^2 =3249 3249 =57
5858 ^2 =3364 3364 =58
5959 ^2 =3481 3481 =59
6060 ^2 =3600 3600 =60
6161 ^2 =3721 3721 =61
6262 ^2 =3844 3844 =62
6363 ^2 =3969 3969 =63
6464 ^2 =4096 4096 =64
6565 ^2 =4225 4225 =65
6666 ^2 =4356 4356 =66
6767 ^2 =4489 4489 =67
6868 ^2 =4624 4624 =68
6969 ^2 =4761 4761 =69
7070 ^2 =4900 4900 =70
7171 ^2 =5041 5041 =71
7272 ^2 =5184 5184 =72
7373 ^2 =5329 5329 =73
7474 ^2 =5476 5476 =74
7575 ^2 =5625 5625 =75
7676 ^2 =5776 5776 =76
7777 ^2 =5929 5929 =77
7878 ^2 =6084 6084 =78
7979 ^2 =6241 6241 =79
8080 ^2 =6400 6400 =80
8181 ^2 =6561 6561 =81
8282 ^2 =6724 6724 =82
8383 ^2 =6889 6889 =83
8484 ^2 =7056 7056 =84
8585 ^2 =7225 7225 =85
8686 ^2 =7396 7396 =86
8787 ^2 =7569 7569 =87
8888 ^2 =7744 7744 =88
8989 ^2 =7921 7921 =89
9090 ^2 =8100 8100 =90
9191 ^2 =8281 8281 =91
9292 ^2 =8464 8464 =92
9393 ^2 =8649 8649 =93
9494 ^2 =8836 8836 =94
9595 ^2 =9025 9025 =95
9696 ^2 =9216 9216 =96
9797 ^2 =9409 9409 =97
9898 ^2 =9604 9604 =98
9999 ^2 =9801 9801 =99
100100 ^2 =10000 10000 =100

Say we have study of two categorical variables each with only two levels. One of the response levels is considered the 'success' response and the other the 'failure' response. A general 2 × 2 table of the observed counts would be as follows:

Success Failure Total

Group 1

A

B

A + B

Group 2

C

D

C + D

The observed counts in this table represent the following proportions:

Success Failure Total

Group 1

(hat{p}_1=frac{A}{A+B})

(1-hat{p}_1)

A + B

Group 2

(hat{p}_2=frac{C}{C+D})

(1-hat{p}_2)

C + D

Page

Recall from our Z-test of two proportions that our null hypothesis is that the two population proportions, (p_1) and (p_2), were assumed equal while the two-sided alternative hypothesis was that they were not equal.

This null hypothesis would be analogous to the two groups being independent.

Also, if the two success proportions are equal, then the two failure proportions would also be equal. Note as well that with our Z-test the conditions were that the number of successes and failures for each group was at least 5. That equates to the Chi-square conditions that all expected cells in a 2 × 2 table be at least 5. (Remember at least 80% of all cells need an expected count of at least 5. With 80% of 4 equal to 3.2 this means all four cells must satisfy the condition).

When we run a Chi-square test of independence on a 2 × 2 table, the resulting Chi-square test statistic would be equal to the square of the Z-test statistic (i.e., ((Z^*)^2)) from the Z-test of two independent proportions.

Political Affiliation and Opinion Section

Consider the following example where we form a 2 × 2 for the Political Party and Opinion by only considering the Favor and Opposed responses:

favoropposeTotal

democrat

138

64

202

republican

64

84

148

Total

202

148

350

The Chi-square test produces a test statistic of 22.00 with p-value 0.00

The Z-test comparing the two sample proportions of (hat{p}_d=frac{138}{202}=0.683) minus (hat{p}_r=frac{64}{148}=0.432) results in a Z-test statistic of (4.69) with p-value of (0.000).

If we square the Z-test statistic, we get (4.69^2 = 21.99) or (22.00) with rounding error.

Try it! Section

The condiments and gender data was condensed to consider gender and either mustard or ketchup. The manager wants to know if the proportion of males that prefer ketchup is the same as the proportion of females that prefer ketchup. Test the hypothesis two ways (1) using the Chi-square test and (2) using the z-test for independence with significance level of 10%. Show how the two test statistics are related and compare the p-values.

Condiment
GenderKetchupMustardTotal
Male152338
Female251944
Total404282

Z-test for two proportions

The hypotheses are:

(H_0colon p_1-p_2=0)

(H_acolon p_1-p_2ne 0)

Let males be denoted as sample one and females as sample two. Using the table, we have:

(n_1=38) and (hat{p}_1=frac{15}{38}=0.395)

(n_2=44) and (hat{p}_2=frac{25}{44}=0.568)

The conditions are satisfied for this test (verify for extra practice).

To calculate the test statistic, we need:

(p^*=dfrac{x_1+x_2}{n_1+n_2}=dfrac{15+25}{38+44}=dfrac{40}{82}=0.4878)

The test statistic is:

begin{align} z^*&=dfrac{hat{p}_1-hat{p}_2-0}{sqrt{p^*(1-p^*)left(frac{1}{n_1}+frac{1}{n_2}right)}}&=dfrac{0.395-0.568}{sqrt{0.4878(1-0.4878)left(frac{1}{38}+frac{1}{44}right)}}&=-1.567end{align}

Numbers

The p-value is (2P(Z<-1.567)=0.1172).

The p-value is greater than our significance level. Therefore, there is not enough evidence in the data to suggest that the proportion of males that prefer ketchup is different than the proportion of females that prefer ketchup.

Chi-square Test for independence

The expected count table is:

Condiment
GenderKetchupMustardTotal
Male15 (18.537)23 (19.463)38
Female25 (21.463)19 (22.537)44
Total404282

There are no expected counts less than 5. The test statistic is:

(chi^{2*}=dfrac{(15-18.537)^2}{18.537}+dfrac{(23-19.463)^2}{19.463}+dfrac{(25-21.463)^2}{21.463}+dfrac{(19-22.537)^2}{22.537}=2.46 )

With 1 degree of freedom, the p-value is 0.1168. The p-value is greater than our significance value. Therefore, there is not enough evidence to suggest that gender and condiments (ketchup or mustard) are related.

Squares

Comparison

The p-values would be the same without rounding errors (0.1172 vs 0.1168). The z-statistic is -1.567. The square of this value is 2.455 which is what we have (rounded) for the chi-square statistic. The conclusions are the same.