Question

Gallup recently conducted a survey about the proportion of Americans who use e-cigarettes (vape). The following example is based on hypothetical data that is meant to be similar to the data collected by Gallup.

The following table considers survey data about annual household income and whether or not a person vapes.

	<$35,000	$35,000-$99,999	$100,000+	Total
Vape	47	57	19	123
Do Not Vape	381	659	362	1402
Total	428	716	381	1525

Problem 1. Based on recent data, about 28% of Americans earn less than $35,000 annually, about 42% of Americans earn between $35,000 and $99,999 annually, and about 30% of Americans earn more than $100,000. Does it appear that the sample is representative of the population? In other words, does it appear that the total people in each income category matches the appropriate proportion? Conduct a chi-square goodness of fit test at the 5% significance level by completing the following steps:

1. Compute the test statistic using the observed frequencies from the table and the expected frequencies you computed in part (b).
2. state the degrees of freedom and find the critical value.
3. Answer the question: does it appear that the sample is representative of the population? Justify using either the critical value method or p-value method.

Answer 1

1.

Given table data is as below MATRIX col1 col2 col3 TOTALS row 1 47 57 19 123 row 2 381 659 362 1402 TOTALS 428 716 381 N = 1525 ------------------------------------------------------------------ calculation formula for E table matrix E-TABLE col1 col2 col3 row 1 row1*col1/N row1*col2/N row1*col3/N row 2 row2*col1/N row2*col2/N row2*col3/N ------------------------------------------------------------------ expected frequencies calculated by applying E - table matrix formula E-TABLE col1 col2 col3 row 1 34.521 57.75 30.73 row 2 393.479 658.25 350.27 ------------------------------------------------------------------ calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 47 34.521 12.479 155.725 4.511 57 57.75 -0.75 0.563 0.01 19 30.73 -11.73 137.593 4.477 381 393.479 -12.479 155.725 0.396 659 658.25 0.75 0.563 0.001 362 350.27 11.73 137.593 0.393 ᴪ^2 o = 9.788 ------------------------------------------------------------------ set up null vs alternative as null, Ho: no relation b/w X and Y OR X and Y are independent alternative, H1: exists a relation b/w X and Y OR X and Y are dependent level of significance, α = 0.05 from standard normal table, chi square value at right tailed, ᴪ^2 α/2 =5.991 since our test is right tailed,reject Ho when ᴪ^2 o > 5.991 we use test statistic ᴪ^2 o = Σ(Oi-Ei)^2/Ei from the table , ᴪ^2 o = 9.788 critical value the value of |ᴪ^2 α| at los 0.05 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 3 - 1 ) = 1 * 2 = 2 is 5.991 we got | ᴪ^2| =9.788 & | ᴪ^2 α | =5.991 make decision hence value of | ᴪ^2 o | > | ᴪ^2 α| and here we reject Ho ᴪ^2 p_value =0.007 ANSWERS --------------- null, Ho: no relation b/w X and Y OR X and Y are independent alternative, H1: exists a relation b/w X and Y OR X and Y are dependent a. test statistic: 9.788 b. critical value: 5.991 c. p-value:0.007 decision: reject Ho we have enough evidence to support the claim that it appear that the sample is representative of the population.