Chi-Square-Test

  1. A local area homeless shelter wanted to assess the impact of a new outreach program aimed at getting homeless people off of the streets at night during extremely cold weather. On a very cold night the homeless shelter sent out two groups of workers to try and get homeless people to stay overnight at the shelter. One group of workers used the new outreach intervention while the other group used the current standard practice. The data from the experiment are provided in the table below. 30 people were approached by the outreach group and 45 were approached by the standard practice group. You want to determine if the new outreach program had an impact on the proportion of homeless people staying at the shelter. This test should be done while keeping the probability of incorrectly rejecting the null hypothesis, when it is fact true, to be less than 5 in 100.

  2.       

    a) What is the appropriate statistical test to use?

    Ans : Chi-square test.

    b) What should the alpha level for this test be?

    Ans: 0.05

    c) Test the null hypothesis of no relationship between theuse of the new outreach intervention and sleeping at the homeless shelter.Write the null and alternative hypotheses, perform the statistical test and write your conclusion?

    Ans:Pearson's Chi-squared test with Yates' continuity correction
    data:  chi_data

    X-squared = 5.0625, df =  1,  p-value =  0.02445
    Since the p-values 0.02445, which is less than 0.05, hence we reject the null hypothesis and conclude there is a relationship between the use of the new outreach intervention and sleeping at the homeless shelter.

  3. This problem requires a 2-way chi-square test, which you will do by hand. In the table below, you see the data from a marketing study in which men and women were asked their preferences for three types of hearing aids, which we will call X,Y, and Z. We want to test whether the frequencies of these choices were significantly different from what we would expect by chance, i.e. what we would expect from purely random choices.

           


    Ans :

          

    DF = (r - 1) * (c - 1) = (2 - 1) * (3 - 1) = 2

    Er,c = (nr * nc) / n

       Expexted Value

    E1,1  =42*26/77  =14.182
    E1,2  =42*26/77  =14.182
    E1,3  =42*25/77  =13.636
    E1,4  =35*26/77  =11.818
    E1,5  =35*26/77  =11.818
    E1,6  =35*25/77  =11.364

    Chi-square = Χ2 = Σ [ (Or,c - Er,c)2 / Er,c ]

          

    Chi-square  = 4.642

    At 2 degrees of freedom and 95% confidence interval the critical value from table is 0.103
    Since the calculated chi-square value 4.642 is greater than 0.103 , we can conclude that the frequencies of their choices are not significantly different.

  4. Arandom sample of 150 employees hired last year showed the relationships in the table betwiin number of working ear and promotion. Use 5% significance level.
    Using the attached data answer the following.

    1) What is your hypothesis testing.

    2) What is your x2.

    3) Is the x2 value significant at 5% significance level.

    4) Write the conclusion for this question.

    PLEASE SHOW ALL WORK PREFERABLY ON PAPER LIST ALL STEPS PLEASE

  5.       

    Ans.

    1) Null Hypothesis : Working years and promotions are independent.
    Alternative Hypothesis: Working years and promotions are significantly different from each other.

    2) Test statistic : Chi-square calculation
    Degrees of freedom = (no. of rows -1) * (no of coloumns - 1) = (2-1)*(2-1) = 1

            Expected
    E1,1      =30*25/150 = 5
    E1,2      =120*25/150 = 20
    E2,1      =30*125/150 = 25
    E2,2      =120*125/150 = 100

    Formula:X2=Σ [ (Oi - Ei)2 / Ei ]

    Expected(E)  Observed(O)  X2
      5     9     3.2
      20    16     0.8
      25    21     0.64
      100   104     0.16
          Total     4.8

    Chi square = 4.8

    3) The significant value at 5% with 1 degrees of freedom is 0.004

    4) Conclusion : Since the calculated Chi-square value is 4.8 which is greater than ssignificant level 0.004, hence we reject the null hypothesis and conclude that Working years and promotions are significantly different from each other.

  6. The sample data below represent the number of late and on tie flights for Delta, United Airlines and US Airways.(Bureau of Trans. Stat. 2012)
          

    A) Formulate the hypotheses for a test thar will determine ifthe population proportion of late flights is the same for all three airlines.
    B) Conduct the hypotheses test with .05 significance level. What is the p value and what is your conclusion?
  7. Ans.

    A) Null Hypothesis: The population proportion is of late flights is the same for all three airlines.
    Alternative Hypothesis : Their is a significant difference between late fights for three sirlines.

    B) Test Statistics:
    Degree of freedom(DF) = K-1
    DF =
    X2=Σ [ (Oi - Ei)2 / Ei ]
    Oi is observed value , Ei is the expected value
    Expected E
    =300*146/1000
    =300*146/1000
    =400*146/1000
    =300*854/1000
    =400*854/1000

          

    Chi-square = 2.12
    The critical value at 0.05 for (3-1)*(2-1) = 2 degrees of freedom is 5.911
    P-value = 2.117
    Conclusion: Since the p-value is greater that 0.05 , hence we accept null hypotheiss and conclude population proportion is of late flights is the same for all three airlines.

  8. In what ways do advertisers in magazines use sexual imagery to appeal to youth? One study classified each of 1509 full-page or larger ads as "not sexual" or "sexual," according to the amount and style of the dress of the male or female model in the ad. The ads were also classified according to the target readership of the magazine. Here is the two-way table of counts.

  9.       

    (a) Summarize the data numerically and graphically. (Compute the conditional distribution of model dress for each audience. Round your answers to three decimal places.)

          

    (b) Perform the significance test that compares the model dress for the three categories of magazine readership. . (Use α = 0.01. Round your value for χ2 to two decimal places, and round your P-value to four decimal places.)

          

    Conclusion

    a) Fail to reject the null hypothesis. There is significant evidence of an association between target audience and model dress
    b) Reject the null hypothesis. There is not significant evidence of an association between target audience and model dress.
    c) Fail to reject the null hypothesis. There is not significant evidence of an association between target audience and model dress.
    d) Reject the null hypothesis. There is significant evidence of an association between target audience and model dress.

    Solution:

    a) Calculate the expected value from the table using the formula - Er,c=(nr*nc)/n
          

          


    b) DF = (r - 1) * (c - 1) = (2 - 1) * (3 - 1) = 2
    Χ2 = Σ [ (Oi - Ei)2 / Ei ]

          

    X2 = 97.150
    P-value = 0.000
    Reject the null hypothesis. There is significant evidence of an association between target audience and model dress.