Question

the total number of horses is 50 and 24 of theses horses required surgery

STA 201 Question 3 (12 marks) In this question, the dataset horse.csv that you have used in Assignment 1 will again be used. You may wish to revisit Assignment 1 for the details regarding the dataset. Based on her experience, a vet suggested that less than half of the horses that show colic symptoms require surgery. At the 5% level of significance, do the data support her statement? (a) Calculate a sample statistic that would best estimate the population (or true) proportion of horses that require surgery. (1 mark] (b) Based on the information provided, write down the null and alternative hypotheses. [1 mark] (c) Write down the name of the test you are going to use. [1 mark] (d) Write down the decision rule. [1 mark] (e) Paste the output from R Commander. (Alternatively, if you would like to do it by hand. calculate the test statistic and show the steps.) [1 mark] (f) Do you reject the null hypothesis or not? Explain your answer. [1 mark] (g) Draw a conclusion which answers the original problem. [1 mark] (h) State and check all the assumptions for the hypothesis test you have used. (3 marks (1) Manually calculate a 90% confidence interval for the population proportion of horses that [3 marks] require surgery. Note: Parts (b) to (8) are actually the six steps of hypothesis testing. You may use it as a guideline for some of the remaining questions. Marking Criteria: 3 0 mark mark Incorrect Correct answer/explanation answer/explanation or not attempted O mark mark 2 marks 3 marks All sumptions Some of the All assumptions All assumptions incorrect or not assumptions are correctly stated but correctly stated and attempted correctly stated and not all correctly correctly checked checked checked O mark 1 mark Final Incorrect final Correct final answer answer answer or not attempted ---- -- marks mark Correct working Process Irrelevant or totally Attempt to do the incorrect method of process but impaired manual calculation not attempted by a minor mistake given

n = 50

number of horses needing surgery = 24

Answer 1

3.

Estimate of population proportion of horses that requires surgery is given by 24 = 0.48 (b) We have to test for null hypothesis Ho: p = 0.5 against the alternative hypothesis H1 :p < 0.5 (c) We have to perform test of proportion for one population case. (d) We have, level of significance a = 0.05 Our decision rule is "we reject our null hypothesis if p-value <0".

(e) Our test statistic is given by - po po (1 - po) /n Here, p = 0.48 po = 0.5 Sample size n = 50 ulated 0.48 – 0.5 == -0.2828427 0.5(1 – 0.5) /50 p-value = P(:< -0.2828427) = 0.3886487 [Using R-code 'pnorm(-0.2828427)]

The R calculation is as follows. >n<-50 > r<-24 > p_hat<-r/n > P_O<-0.5 > Z<-(p_hat-p_0)/sqrt (p_0* (1-p_0)/n) > Z [1] -0.2828427 > p<-pnorm (z) 2 [1] 0.3886487 We cannot reject our null hypothesis. We reject our null hypothesis if p-value <a But here we observe that p-value = 0.3886487 0.05 = a

(g) Based on the given data we can conclude that there is no significant evidence that less than half of the horses that show colic symptoms require surgery. (h) We used normal approximation of binomial distribution which is valid if • Number of success 10 • Number of failure 10 Considering having surgery as success we have • Number of success = 24 > 10 • Number of failure = 26 > 10 So, assumption for the hypothesis test is satisfied.

We know, P(-1.644854<:<1.644854) = 0.90 [Using R-code 'qnorm(1-1-0.90)/2)] + P(-1.691851 < p ="BTH < 1.61854) = 0,00 =P(-1.641854 < 70.8415 0.)750 < 1.611854) = 0,90 = P(-1.644854 < 0.04865.205 < 1.644854) = 0.90 - 1 54 <- 0.48 - P 0.48 (1 – 0.48)/50 .644 1 = 0.90 →P(-1.644854 * 0.07065409 < 0.48-P < 1.644854 * 0.07065409) = 0.90 =P(0.48 – 0.1162157 < p < 0.48 +0.1162157) = 0.90 :P(0.3637843<p < 0.5962157) = 0.90 Hence, our calculated 90% confidence interval for the population proportion of horses that require surgery is given by (0.3637843, 0.5962157).