Question

In: Statistics and Probability

A survey of 100,000 college graduates and 100,000 non-college graduates is conducted to determine how many...

A survey of 100,000 college graduates and 100,000 non-college graduates is conducted to determine how many of them own stock. 49,781 of the non-college graduates own stock vs. 50,219 for the college graduates. Analyze for a relationship between education and owning stock using a chi-square hypothesis test with alpha=1%.

Solutions

Expert Solution

Data Summary

Observed Frequencies (O) College Non-college
Own stock 50219 49781
Do not own stock 49781 50219

The null and alternative hypotheses are   
Ho : Education and Owning stock are independent   
Ha : Education and Owning stock are dependent   

   
Level of significance α = 0.01   
We use Chi Square test of independence   
   
Grand Total of frequencies = 200000   
To find Expected Frequencies   
Expected Frequencies = (Row Total * Column Total)/Grand Total   

Expected Frequencies (E) College Non-college
Own stock 50000 50000
Do not own stock 50000 50000

Following table gives the value of (Observed - Expected)² / Expected

College Non-college
Own stock 0.9592 0.9592
Do not own stock 0.9592 0.9592

Chi Square Value = ∑[(Observed - Expected)² / Expected]   
χ² = 3.8369   
   
Degrees of freedom = df = (number of rows - 1) * (number of columns - 1)   
Degrees of freedom = df = 1   
   
From chi square table or CHISQ.DIST.RT(X, df) function of Excel   
we find the p-value   
p-value = CHISQ.DIST.RT( 3.8369, 1) = 0.0501   
p-value = 0.0501   

Decision :   
0.0501 > 0.01   
that is p-value > α   
Hence we DO NOT REJECT Ho   
      
Conclusion :   
There does not exist enough statistical evidence at α = 0.01 to show that Education and Owning stock are dependent.

that is

There is no relationship between Education and Owning stock.   


Related Solutions

A survey was conducted at a college to determine the number of hours students spent outside...
A survey was conducted at a college to determine the number of hours students spent outside on weekends during the winter. For students from cold weather climates, the results were M = 8, SD = 1.5. For students from warm-weather climates, the results were M = 5, SD = 4.2. Explain what these numbers mean to a person who has never had a course in statistics, and suggest conclusions the person can draw from these results. These conclusions use the...
A recent report claims that non-college graduates get married at an earlier age than college graduates....
A recent report claims that non-college graduates get married at an earlier age than college graduates. To support the claim, random samples of size 100 were selected from each group, and the mean age at the time of marriage was recorded. The mean and standard deviation of the non-college graduates were 22.5 years and 1.4 years respectively, while the mean and standard deviation of college graduates were 23 years and 1.8 years respectively. Test the hypothesis (the claim being made)...
A recent survey was conducted to determine how people consume their news. According to this​ survey,...
A recent survey was conducted to determine how people consume their news. According to this​ survey, 60​% of men preferred getting their news from television. The survey also indicates that 42​% of the sample consisted of males. ​ Also, 42​% of females prefer getting their news from television. Use this information to answer the following questions. a. Consider that you have 30000 ​people, given the information​ above, how many of them are​ male? b. Again considering that you have 30000...
The figure to the right shows the results of a survey in which 2500 college graduates...
The figure to the right shows the results of a survey in which 2500 college graduates from the year 2016 were asked questions about employment. Construct 95​% confidence intervals for the population proportion of college students who gave each response. A table labeled "Employment, College students' responses to questions about employment" consists of five rows containing the following information from top to bottom, with row listed first and information listed second: Expect to stay at first employer for 3 or...
A survey was conducted that asked 1008 people how many books they had read in the...
A survey was conducted that asked 1008 people how many books they had read in the past year. Results indicated that x overbarequals14.4 books and sequals16.6 books. Construct a 99​% confidence interval for the mean number of books people read. Interpret the interval.
A survey was conducted that asked 1008 people how many books they had read in the...
A survey was conducted that asked 1008 people how many books they had read in the past year. Results indicated that x overbarequals14.4 books and sequals16.6 books. Construct a 99​% confidence interval for the mean number of books people read. Interpret the interval.
A survey was conducted that asked 1023 people how many books they had read in the...
A survey was conducted that asked 1023 people how many books they had read in the past year. Results indicated that x over bar x equals=15.4 books and s equals=18.2 books. Construct a 95​% confidence interval for the mean number of books people read. Interpret the interval. confidence interval for the mean number of books people read and interpret the result. Select the correct choice below and fill in the answer boxes to complete your choice. ​(Use ascending order. Round...
A survey was conducted that asked 1019 people how many books they had read in the...
A survey was conducted that asked 1019 people how many books they had read in the past year. Results indicated that x overbarequals10.9 books and sequals16.6 books. Construct a 99​% confidence interval for the mean number of books people read. Interpret the interval. Construct a 99​% confidence interval for the mean number of books people read and interpret the result. Select the correct choice below and fill in the answer boxes to complete your choice. ​(Use ascending order. Round to...
A survey was conducted that asked 1016 people how many books they had read in the...
A survey was conducted that asked 1016 people how many books they had read in the past year. Results indicated that x overbar = 14.9 books and s = 16.6 books. Construct a 90​% confidence interval for the mean number of books people read. Interpret the interval.
A survey was conducted that asked 1014 people how many books they had read in the...
A survey was conducted that asked 1014 people how many books they had read in the past year. Results indicated that x overbarequals14.7 books and sequals16.6 books. Construct a 95​% confidence interval for the mean number of books people read. Interpret the interval. LOADING... Click the icon to view the table of critical​ t-values. Construct a 95​% confidence interval for the mean number of books people read and interpret the result. Select the correct choice below and fill in the...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT