Question

It is known from past information that the probability of failing to finish an Ironman (called a DNF) is 15%. Suppose we take a sample of 100 Ironman races over the past few years and find that the DNF percentage is 12%. We are interested in seeing if there has been a significant decrease in the number of DNFs. Use alpha=0.05. Find the 95% confidence interval for the percentage of DNFs.

	a.	(0.0665, 0.1735)
	b.	(0.0555, 0.1845)
	c.	(0.1179, 0.1221)
	d.	(0.0563, 0.1837)

It is known from past information that the probability of failing to finish an Ironman (called a DNF) is 15%. Suppose we take a sample of 100 Ironman races over the past few years and find that the DNF percentage is 12%. We are interested in seeing if there has been a significant decrease in the number of DNFs. Use alpha=0.05. The null and the alternative hypotheses are:

	a.	Null hypothesis: The population proportion is less than or equal to 0.12 Alternative hypothesis: The population proportion is greater than 0.12
	b.	Null hypothesis: The population proportion is less than or equal to 0.15 Alternative hypothesis: The population proportion is greater than 0.15
	c.	Null hypothesis: The population proportion is greater than or equal to 0.15 Alternative hypothesis: The population proportion is less than 0.15
	d.	Null hypothesis: The sample proportion is less than or equal to 0.15 Alternative hypothesis: The sample proportion is greater than 0.15

It is known from past information that the probability of failing to finish an Ironman (called a DNF) is 15%. Suppose we take a sample of 100 Ironman races over the past few years and find that the DNF percentage is 12%. We are interested in seeing if there has been a significant decrease in the number of DNFs. Use alpha=0.05. The test statistic is:

	a.	t= -0.84
	b.	t= -0.92
	c.	z= -0.84
	d.	z= -0.92

It is known from past information that the probability of failing to finish an Ironman (called a DNF) is 15%. Suppose we take a sample of 100 Ironman races over the past few years and find that the DNF percentage is 12%. We are interested in seeing if there has been a significant decrease in the number of DNFs. Use alpha=0.05. The p-value is:

	a.	0.1798
	b.	0.2005
	c.	0.7995
	d.	p-value greater than 0.20

It is known from past information that the probability of failing to finish an Ironman (called a DNF) is 15%. Suppose we take a sample of 100 Ironman races over the past few years and find that the DNF percentage is 12%. We are interested in seeing if there has been a significant decrease in the number of DNFs. Use alpha=0.05. Assuming no prior estimate of p is available, what is the sample size that would need to be taken if we wanted to have a margin of error of 0.08 or less with 95% confidence?

	a.	106
	b.	150
	c.	150.063
	d.	151

Answer 1

1)

Level of Significance,   α =    0.05          
Number of Items of Interest,   x =   12          
Sample Size,   n =    100          
                  
Sample Proportion ,    p̂ = x/n =    0.120          
z -value =   Zα/2 =    1.960   [excel formula =NORMSINV(α/2)]      
                  
Standard Error ,    SE = √[p̂(1-p̂)/n] =    0.0325          
margin of error , E = Z*SE =    1.960   *   0.0325   =   0.0637
                  
95%   Confidence Interval is              
Interval Lower Limit = p̂ - E =    0.120   -   0.0637   =   0.0563
Interval Upper Limit = p̂ + E =   0.120   +   0.0637   =   0.1837
                  
answer:

d.(0.0563, 0.1837)

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2)

Null hypothesis: The population proportion is greater than or equal to 0.15

Alternative hypothesis: The population proportion is less than 0.15

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3)

Sample Proportion ,    p̂ = x/n =    0.120                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.0357                  
Z Test Statistic = ( p̂-p)/SE = (   0.1200   -   0.15   ) /   0.0357   =   -0.84

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4)

without prior estimate, let    sample proportion ,   p̂ =    0.5                          
   sampling error ,    E =   0.08                          
   Confidence Level ,   CL=   0.95                          
                                      
   alpha =   1-CL =   0.05                          
   Z value =    Zα/2 =    1.960   [excel formula =normsinv(α/2)]                      
                                      
   Sample Size,n = (Z / E)² * p̂ * (1-p̂) = (   1.960   /   0.08   ) ² *   0.50   * ( 1 -   0.50   ) =    150.1
                                      
                                      
   so,Sample Size required=       151