Question

PART B: Written Questions. Please answer both questions, (40% each) 1. A large Irish sample survey asked each individual to describe their own health levels using five alternative categories. The numbers falling into each category were as follows: Self-Described Frequency Health Status Very Good Good 3270 3376 1260 244 air Bad Very Bad Total 51 8201 Calculate the proportion of the sample describing their health as either 'bad' or 'very bad' a construct a 95% confidence interval for the population proportion with bad or very bad health Explain clearly how this confidence interval should be interpreted. Comment on the margin of (a) error. In the same survey, it was found that, for the 51 i income was 18,647, with a sample standard deviation of 10,519. Mean income for the whole Irish population in the year of the survey was 23,334. You are asked to establish whether the mean income of individuals in very bad health differed from 23,344. Carry out an appropriate hypothesis test, using a 95% confidence level. (b) ndividuals with very bad health, their mean Using a diagram, explain how using a lower confidence level of 90% for parts (a) and (b) above would affect your answers.

Answer 1

Answer:

a)

The sample proportion is 244+51 51.0360 P8201 Here we have x295, n-8201,p0.036 Level of significance: a-0.05 Critical value: 1.96 The standard deviation is: P-)sqrt(0.036*(1-0.036)/8201) -0.0021 The requried confidence interval is: (1.96*0.0021) -0.036 ptz,.-, (0.032,0.04 Excel function for critical value: NORMSINV(1-(0.05/2)) Hence, the required confidence interval is (0.032,0.04).

b)

Here we have r-18647, s -10519, n -51 Hypotheses are: Но, щ-23334 Ha: μ# 23334 Level of significance: 0.05 Test is two tailed. Degree of freedom: df -n-1-50 Test Statistics: - (18647-23334)/(10519/sqrt(51) -3.182 Critical value: 2.009 If It >2.009, Reject HO Rejection Region: Decision: Since test statistics lies in rejection region so we reject the null hypothesis P-value: P-value-0.0025 Decision: Since p-value is less than level of significance so we reject the null hypothesis Excel function for critical value: TINV(0.05,50) Excel function for p-value -TDIST(3.182,50,2)

c)

As the confidence level decrease width of confidence interval also decreases And as the confidence level decreases level of significance increases so chances to rejecting the null hypothesis increases