Question

In 2013, according to the American Dental Association, 50% of adults floss their teeth daily. Wondering if the percentage is less today, a recent Gallup poll was conducted and found that 436 out of 960 adults floss daily. Perform a hypothesis test at the ? = 0.05 level of significance. Is there sufficient evidence to conclude that the percentage of Americans who floss today is less than 50%?

1. what is the test statistic?

2. what is the p-value

3. Construct a 95% confidence interval.

4. Do the p-value and confidence interval approaches bring you to the same conclusion? Explain why or why not.

5. State your conclusion in the context of the problem

Answer 1

H0: p = 0.5

Ha: p < 0.5

1)

Sample proportion = 436 / 960 = 0.4542

Test statistics

z = ( - p) / sqrt [ p ( 1 - p) / n ]

= ( 0.4542 - 0.50) / sqrt ( 0.50 * 0.50 / 960 )

= -2.84

2)

p-value = P(Z < z)

= P(Z < -2.84)

= 0.0023

3)

p̂ = X / n = 436/960 = 0.454
p̂ ± Z(α/2) √( (p * q) / n)
0.4542 ± Z(0.05/2) √( (0.4542 * 0.5458) / 960)
Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.454 - Z(0.05) √( (0.4542 * 0.5458) / 960) = 0.423
upper Limit = 0.454 + Z(0.05) √( (0.4542 * 0.5458) / 960) = 0.486
95% Confidence interval is ( 0.423 , 0.486 )

4)

Using p-value approach,

Since p-value < 0.05 significance level , Reject the null hypothesis.

Using confidence interval approach,

Since 0.50 does not contained in confidence interval and all values in confidence interval

are less than 0.50, reject the null hypothesis.

5)

Conclusion = We have sufficient evidence to support the claim that the percentage of adults

floss their teeth daily is less than 50%