A study was conducted to determine if there is a difference between the investin

Question

A study was conducted to determine if there is a difference between the investing preferences of mid-level managers working in the public and private sectors in New York City. A random sample of 320 public sector employees and 380 private sector employees was taken. The sampled participants were then asked about their retirement investment decisions and classified as being either “aggressive,” if they invested only in stocks or stock mutual funds, or “balanced,” if they invested in some combination of stocks, bonds, cash, and other. The following results were found: Agressive Balanced Public 164 156 Private 236 144 a. State the hypothesis of interest and conduct the appropriate hypothesis test to determine whether there is a relationship between employment sector and investing preference. Use a level of significance of 0.01. (3 marks) b. State the conclusion of the test conducted in part a. (3 marks) c. Calculate the p-value for the hypothesis test conducted in part a. (4 marks)

Explanation / Answer

(a ) We are requirred to state hypotheses and conduct a test to see if there is any relation between employment sector and innestinf preference The hypothesis are defined as H0: There is no relation between employment sector and investing preference H1: There is relation between employment sector and the inesting preference We will do chi-square test. If null hypothesis is true the expected cell frequencies are calculated as (row total*Column total)/table total Corresponding to the row and column of the cell the expected cell frequencies are calculated as Aggressive Balanced Public 182.8571 137.1429 Private 217.1429 162.8571 if the null hypothesis is true value of Chi-square statistis is calculated as (Observed Frequency-expected frequency)^2/expected frequency =(164-182.8571)^2/182.8571+(156-137.1429)^2/137.1429+(236-217.1429)^2/217.1429+(144-162.8571)^2/162.8571      8.3585 b The degree of freedom is (row-1)*(colum-1)=(2-1)*(2-1)=1 The Critical value of chi squre with 1 degree of freedom and level of significance =0.1 is 6.63 since the calculated value of chi suare (8.36) is greater than the critical value of chi-square at 0.01 level of significance (6.63) we reject the null hypothesis c P value of the hypothesis =CHITEST(K7:L8,K22:L23) 0.003839 The p-value is .003839

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