Question

Please fill in the following table based on the results for the entire class. Head Tail Total Penny 19 11 30 Nickel 14 16 30 Dime 6 4 10 Quarter 16 14 30 Total 100 1. What is the proportion of coins that landed on a head? 2. What is the proportion of nickels that landed on a tail? 3. What is the proportion of heads that were dimes? 4. What proportion of the tosses were pennies and tails? 5. What is the sample proportion of quarters that landed on heads? 6. What is the 95% confidence interval for the true proportion of quarters that would land on heads? Interpret your interval! 7. Are the quarters fair? Perform a hypothesis test with a significance level of 0.05. State your hypotheses, test statistic, P-value, and conclusion in terms of the question. 8. Could you have used the confidence interval from #5 to perform the test in #6? Why or why not? 9. Is there a significant association with the type of coin and which side the coin lands on? Perform a chi-square test to answer this question. State your hypotheses, test statistic, P-value, and conclusion in terms of the question.

Answer 1

Type	Head	Tail	Total
Penny	19	11	30
Nickel	14	16	30
Dime	6	4	10
Quarter	16	14	30
Total	55	45	100

The above table denotes observed frequency

1. The proportion of coins that landed in a head is 55C1/100C1= 55/100=0.55

2. The proportion of nickel that landed on a tail= 16C1/30C1=16/30=0.533

3. The proportion of head that are dime= 6C1/10C1=6/10=0.6

4. The pennies be denoted as "A" and tails as "B" we have to find P(AB)= 30/100*45/100=.3*.45=.135

5. The sample proportion of quarters that landed on head=16C1/30C1=16/30=0.533

6. Standard Error (S.E)= where p=1/2 q=1/2 n =30 therefore S.E=.09128

the 95% confidence limits == (.32109, 0.6789)

7. Ho: The quarters are fair

Ha: The quarters are not fair

IzI = (Observed proportion- Expected proportion)/ S.E where observed proportion =16/30=.533 expected proportion=0.5 and S.E=.09128

Therefore IzI=0.33 with p value 0.251. We are performing the test at 5% level of significance. So calculated p value is greater than 0.05 hance null hypothesis is accepted i.e. the quarters are probably fair

8. If we have calculated S.E by sample proportions with p=16/30 and q=14/30 then S.E=.09108. We cannot use it since there will be error creeping in. Here the error of S.E=(.09128-.09108)=.0002 which will result in error of confidence interval.

9. H0: There is no association with the type of coin and which side it landed

Ha: There is association with the type of coin and which side it landed

Expected Frequency table

Type	Head	Tail	Total
Penny	15	15	30
Nickel	15	15	30
Dime	5	5	10
Quarter	15	15	30
Total	50	50	100

= (O-E)2/E=2.53 where O= Observed frequency and E=Expected Frequency.

The degrees of freedom is (4-1)*(2-1)=3 and we are testing at 5% level of significance.The calculated value is 2.53 with p value 0.469894 which is greater than 0.05 So the null hypothesis is accepted i.e. there is no association with the type of coin and which side it landed