<< Chapter < Page Chapter >> Page >

The table contains observed ( O ) frequency values.

At the 1% significance level, does it appear that the distribution "number of televisions" of far western United States families is different from the distribution for the Americanpopulation as a whole?

This problem asks you to test whether the far western United States families distribution fits the distribution of the American families. This test is always right-tailed.

The first table contains expected percentages. To get expected ( E ) frequencies, multiply the percentage by 600. The expected frequencies are:

Number of Televisions Percent Expected Frequency
0 10 ( 0.10 ) ( 600 ) = 60
1 16 ( 0.16 ) ( 600 ) = 96
2 55 ( 0.55 ) ( 600 ) = 330
3 11 ( 0.11 ) ( 600 ) = 66
over 3 8 ( 0.08 ) ( 600 ) = 48

Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60,enter .10*600.

H o : The "number of televisions" distribution of far western United States families is the same as the "number of televisions" distribution of the American population.

H a : The "number of televisions" distribution of far western United States families is different from the "number of televisions" distribution of the American population.

Distribution for the test: χ 4 2 where df = (the number of cells) - 1 = 5 - 1 = 4 .

df ≠ 600 − 1

Calculate the test statistic: χ 2 = 29.65

Graph:

Non-symmetric chi-square curve with values of 0, 4, and 29.65 on the x-axis representing the test statistic of the comparison of the number of televisions in America. A vertical upward line extends from 29.65 to the curve, and the area to the right of this line is equal to the p-value.

Probability statement: p-value = P ( χ 2 > 29.65 ) = 0.000006 .

Compare α and the p-value:

  • α = 0.01
  • p-value = 0.000006
So, α > p-value .

Make a decision: Since α > p-value , reject H o .

This means you reject the belief that the distribution for the far western states is the same as that of the American population as a whole.

Conclusion: At the 1% significance level, from the data, there is sufficient evidence to conclude that the "number of televisions" distribution for the far western United Statesis different from the "number of televisions" distribution for the American population as a whole.

TI-83+ and some TI-84 calculators: Press STAT and ENTER . Make sure to clear lists L1 , L2 , and L3 if they have data in them (see the note at the end of Example 11-2). Into L1 , put the observed frequencies 66 , 119 , 349 , 60 , 15 . Into L2 , put the expected frequencies .10*600, .16*600 , .55*600 , .11*600 , .08*600 . Arrow over to list L3 and up to the name area "L3" . Enter (L1-L2)^2/L2 and ENTER . Press 2nd QUIT . Press 2nd LIST and arrow over to MATH . Press 5 . You should see "sum" (Enter L3) . Rounded to 2 decimal places, you should see 29.65 . Press 2nd DISTR . Press 7 or Arrow down to 7:χ2cdf and press ENTER . Enter (29.65,1E99,4) . Rounded to 4 places, you should see 5.77E-6 = .000006 (rounded to 6 decimal places) which is the p-value.

The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF . To run the test, put the observed values (the data) into a first list and the expected values (thevalues you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF . Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw . Make sure you clear any lists before you start.

Suppose you flip two coins 100 times. The results are 20 HH, 27 HT, 30 TH, and 23 TT. Are the coins fair? Test at a 5%significance level.

This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is {HH, HT, TH, TT}. Out of 100 flips, you would expect 25 HH, 25 HT,25 TH, and 25 TT. This is the expected distribution. The question, "Are the coins fair?" is the same as saying, "Does the distribution of the coins (20 HH, 27 HT, 30 TH, 23 TT)fit the expected distribution?"

Random Variable: Let X = the number of heads in one flip of the two coins. X takes on the value 0, 1, 2. (There are 0, 1, or 2 heads in the flip of 2 coins.) Therefore, the number of cells is 3 . Since X = the number of heads, the observed frequencies are 20 (for 2 heads), 57 (for 1 head), and 23 (for 0 heads or both tails). The expectedfrequencies are 25 (for 2 heads), 50 (for 1 head), and 25 (for 0 heads or both tails). This test is right-tailed.

H o : The coins are fair.

H a : The coins are not fair.

Distribution for the test: χ 2 2 where df = 3 - 1 = 2 .

Calculate the test statistic: χ 2 = 2.14

Graph:

Nonsymmetrical chi-square curve with values of 0 and 2.14 on the x-axis representing the test statistic of results from flipping a coin. A vertical upward line extends from 2.14 to the curve and the area to the right of this is equal to the p-value.

Probability statement: p-value = P ( χ 2 > 2.14 ) = 0.3430

Compare α and the p-value:

  • α = 0.05
  • p-value = 0.3430
So, α < p-value .

Make a decision: Since α < p-value , do not reject H o .

Conclusion: There is insufficient evidence to conclude that the coins are not fair.

TI-83+ and some TI- 84 calculators: Press STAT and ENTER . Make sure you clear lists L1 , L2 , and L3 if they have data in them. Into L1 , put the observed frequencies 20 , 57 , 23 . Into L2 , put the expected frequencies 25 , 50 , 25 . Arrow over to list L3 and up to the name area "L3" . Enter (L1-L2)^2/L2 and ENTER . Press 2nd QUIT . Press 2nd LIST and arrow over to MATH . Press 5 . You should see "sum" . Enter L3 . Rounded to 2 decimal places, you should see 2.14 . Press 2nd DISTR . Arrow down to 7:χ2cdf (or press 7 ). Press ENTER . Enter 2.14,1E99,2) . Rounded to 4 places, you should see .3430 which is the p-value.

The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF . To run the test, put the observed values (the data) into a first list and the expected values (thevalues you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF . Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw . Make sure you clear any lists before you start.

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Quantitative information analysis iii. OpenStax CNX. Dec 25, 2009 Download for free at http://cnx.org/content/col11155/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Quantitative information analysis iii' conversation and receive update notifications?

Ask