statistics/

week_6_class_exercises_example.xlsx

Exercise 1

A random sample of 150 runners who finished the New York City Marathon contained 22 who were over 50.

Use a significance level of a = 0.025 to test the claim that fewer than 20% of the NYC marathon runners who

finish are over 50.

A random sample of 150 runners who finished the New York City Marathon contained 22 who were over 50.
Use a significance level of a = 0.025 to test the claim that fewer than 20% of the NYC marathon runners who
finish are over 50.
Claim	p < 0.20
H0	p = 0.20
H1	p < 0.20	left-tailed test		Normal PD with center equal to 0.20
a	0.01			SE = sqrt(p*(1-p)/n)	sqrt(.2*.8/150)
sample statistic		p hat =	0.147
SE	0.0327	Compute the standard error using the proportion in the null hypothesis
p-value	0.05	The probability that a sample statistic is less than the computed value of 0.147
Initial Conclusion		p-value = 0.051 > a = 0.025; therefore, fail to reject the null hypothesis and do not support the claim
Final Conclusion		The sample data do not support the claim that fewer than 20% of the NYC marathon runners who
		finish are over 50.

A random sample of 150 runners who finished the New York City Marathon contained 22 who were over 50.

Test the claim that more than 10% of all finishers are over 50.

A random sample of 150 runners who finished the New York City Marathon contained 22 who were over 50.
Claim that more than 10% of all finishers are over 50.
claim	p > .10
H0	p = .10
H1	p > .10	right tailed
alpha	0.025
sample statistic		0.147
SE	0.0244948974
p-value	0.028
initial conclusion		p-value > alpha, support H0
final conclusion		The sample data do not support the claim that more than 10% of finishers are over 50.

A random sample of 150 runners who finished the New York City Marathon contained 22 who were over 50.

Compute the 95% confidence interval for the population proportion of marathon finishers who are over 50.

Compare the interval to the solutions for Exercises 1 and 2.

A random sample of 150 runners who finished the New York City Marathon contained 22 who were over 50.
Compute the 95% confidence interval for the population proportion of marathon finishers who are over 50.
Compare the interval to the solutions for Exercises 1 and 2.
confidence interval
SE	0.0288854699
P2.5	ERROR:#NAME?
P97.5	ERROR:#NAME?
We are 95% confident that the population proportion of marathon finishers who are over 50 is between 9% and 20.3%.
So cannot claim greater than 10% since could be as small as 9%, and cannot claim less than 20% since could be 20%.

A survey of 205 single women and 260 single men found that 49 of the women and 70 of the men

"definitely want to get married."

Test the claim that the population proportion of women is less than the population proportion of men.

Use a significance level of a = 0.025.

A survey of 205 single women and 260 single men found that 49 of the women and 70 of the men
"definitely want to get married."
Test the claim that the population proportion of women is less than the population proportion of men.
Use a significance level of a = 0.025.
Women		Men
x1	49	x2	70
n1	205	n2	260
p1 hat	0.239	p2 hat	0.269
Claim	p1 < p2			Normal PD with mean zero
H0	p1 = p2
H1	p1 < p2	left tailed
a	0.025
Sample statistic		p1hat - p2 hat	-0.030
p	0.256	pooled proportion
SE	0.041
p-value	0.229
Initial Conclusion		p-value = 0.229 > a = 0.025, therefore fail to reject the null hypothesis and do not support the claim.
Final Conclusion		The sample data do not support the claim that a greater proportion of men "definitely want to get married."

HYPOTHESIS TESTING FOR PROPORTIONS
1. CLAIM concerning a population proportion, will contain <, > , not equal ¹
2. Null Hypothesis		H0	statement of equality, =			p = 0.5
3. Alternative Hypothesis			H1	same as claim
				<	left-tailed test
				>	right-tailed test
				¹	two-tailed		try to avoid
	Normal probability distribution				based on assumption in the Null hypothesis
		bell curve
4. Choose significance level			is the complement of the confidence level
	for example, a confidence level of 95% is equivalent to a significance level of 5%
	confidence level defined the "likely" values
	significance level defines the "unlikely" values, the probability of being in the tail.
5. compute the value of the sample statistic p hat=x/n and the SE
6. Compute probability value or p-value
	p-value = P(sample statistic <= computed sample statistic)						left-tailed test
	p-value = P(sample statistic >= computed sample statistic)						right-tailed test
	p-value = twice the area of the tail defined by p hat						two-taield
7. Initial Conclusion			Reject Null or Do not reject Null/Support Null
			Reject Null and support Alternative=Claim
		or	Do not reject Null and do not support the claim
	rejection rule:		Reject the Null if p-value is less than significance level; otherwise, fail to reject
			the Null.
8. Final Conclusion			Sentence:	The sample data support the claim that….
			or	The sample data do not support the claim that…