Question

1)Background: Morris Saldov conducted a study in Eastern and Central Newfoundland in 1988 to examine public attitudes towards social spending. In particular, the study tried to determine if knowing someone on public assistance (yes, no) affected one's views on social spending (too little, about right, too much). The data from the study is summarized in the table below.

	Yes	No	Total
Too little	38	5	43
About right	15	15	30
Too much	10	5	15
Total	63	25	88

Source: Morris Saldov, Public Attitudes to Social Spending in Newfoundland," Canadian Review of Social Policy, 26, November 1990, pages 10-14.

Directions: Conduct a chi-square test for independence to determine if the association between knowing someone on public assistance and views on social spending is statistically significant.

1. Choose the correct null and alternative hypotheses.
   • H0:H0: There is an association between knowing someone on public assistance and views on social spending.
     HaHa There is no association between knowing someone on public assistance and views on social spending.
   • H0:H0: There is no association between knowing someone on public assistance and views on social spending.
     HaHa There is an association between knowing someone on public assistance and views on social spending.
2. Compute the test statistic.
   
   Complete the following table of expected counts. (Round your answers to 2 decimal places).
   

   	Yes	No
   Too little
   About right
   Too much

   
   Compute the value of the test statistic. (Round your answer to 2 decimal places.)
   
   χ2=χ2=
   
3. Compute the p-value. (Round your answer to 4 decimal places.)
   
   pp-value =
   
4. Interpret the results of the significance test. Use a signifcance level of alpha = 0.05
   • The p-value provides strong evidence against the null hypothesis. The association between knowing someone on public assistance and attitudes towards social spending is statistically significant.
   • The p-value provides little evidence against the null hypothesis. The association between knowing someone on public assistance and attitudes towards social spending is not statistically significant.

2) A financial analyst claims that 19% make all purchases with cash, 17% make most purchases with cash, 20% make half of purchases with cash, 33% make some purchases with cash and 11% make no purchases with cash. You take a random selection to see if you can conclude that the distribution is different than what the financial analyst claims. Use a 5% significance to decide and round to the fourth.

Categories	Observed Frequency	Expected Frequency
All Cash	28
Most Cash	84
Half Cash	103
Some Cash	159
No Cash	94

Test Statistic:
Degrees of Freedom:
p-val:
Decision Rule: Select an answer Reject the Null Accept the Null Fail to Reject the Null
Did something significant happen? Select an answer Nothing Significant Happened Significance Happened
There Select an answer is not is  enough evidence to conclude Select an answer that the distribution is what the financial analyst claims that the distribution is different than what the financial analyst claims

select one - that the is what the financial analyst is claims or is different

Answer 1

H0: There is no association between knowing someone on public assistance and views on social spending.

Ha There is an association between knowing someone on public assistance and views on social spending.

Applying chi square test of independence:

Expected	Ei=row total*column total/grand total	Yes	No	Total
	too little	30.7841	12.2159	43.00
	About right	21.4773	8.5227	30.00
	too much	10.7386	4.2614	15.00
	total	63.00	25.00	88.00
chi square    χ2	=(Oi-Ei)2/Ei	Yes	No	Total
	too little	1.691	4.262	5.9539
	About right	1.953	4.923	6.8762
	too much	0.051	0.128	0.1788
	total	3.6957	9.3132	13.009
test statistic X2 =	13.009

degree of freedom(df) =(rows-1)*(columns-1)=

2

p value =

0.0015

The p-value provides strong evidence against the null hypothesis. The association between knowing someone on public assistance and attitudes towards social spending is statistically significant.

2)

applying chi square goodness of fit test:

	relative	observed	Expected	residual	Chi square
category	frequency(p)	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
1	0.19	28.00	88.92	-6.46	41.737
2	0.17	84.00	79.56	0.50	0.248
3	0.20	103.00	93.60	0.97	0.944
4	0.33	159.00	154.44	0.37	0.135
5	0.11	94.00	51.48	5.93	35.119
total	1.000	468	468		78.1828
test statistic X2 =	78.183

degree of freedom =categories-1=

4

p value =

0.0000

  Significance Happened

There  is  enough evidence to conclude is different than what the financial analyst claims