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Tattoo Attitude A Harris Interactive poll conducted during
Chapter 5, Problem 29AYU(choose chapter or problem)
A Harris Interactive poll conducted during January 2008 found that 944 of 1748 adult Americans 18 years or older who do not have a tattoo believe that individuals with tattoos are more rebellious.
(a) Obtain a point estimate for the proportion of adult Americans without tattoos who believe individuals with tattoos are more rebellious.
(b) Verify that the requirements for constructing a confidence interval for p are satisfied.
(c) Construct a 90% confidence interval for the proportion of adult Americans without tattoos who believe individuals with tattoos are more rebellious.
(d) Construct a 99% confidence interval for the proportion of adult Americans without tattoos who believe individuals with tattoos are more rebellious.
(e) What is the effect of increasing the level of confidence on the width of the interval?
Questions & Answers
QUESTION:
A Harris Interactive poll conducted during January 2008 found that 944 of 1748 adult Americans 18 years or older who do not have a tattoo believe that individuals with tattoos are more rebellious.
(a) Obtain a point estimate for the proportion of adult Americans without tattoos who believe individuals with tattoos are more rebellious.
(b) Verify that the requirements for constructing a confidence interval for p are satisfied.
(c) Construct a 90% confidence interval for the proportion of adult Americans without tattoos who believe individuals with tattoos are more rebellious.
(d) Construct a 99% confidence interval for the proportion of adult Americans without tattoos who believe individuals with tattoos are more rebellious.
(e) What is the effect of increasing the level of confidence on the width of the interval?
ANSWER:
Step 1 of 6
For the given information, a Harris Interactive poll conducted during January 2008 found that 944 of 1748 adult Americans 18 years or older who do not have a tattoo believe that individuals with tattoos are more rebellious.
a) The mean of the distribution of sample proportion is equal to the population proportion.
That is , \(\mu_{\bar{x}}=\widehat{p}\)
Here the given values are x = 944 and n = 1748.