A recent study examined hearing loss data for 1769 U.S. teenagers. In this sampl
ID: 3349789 • Letter: A
Question
A recent study examined hearing loss data for 1769 U.S. teenagers. In this sample, 331 were found to have some level of hearing loss. News of this study spread quickly, with many news articles blaming the prevalence of hearing loss on the higher use of ear buds by teens. At MSNBC.com (8/17/2010), Carla Johnson summarized the study with the headline: “1 in 5 U.S. teens has hearing loss, study says.” To investigate whether this is an appropriate or a misleading headline, you will conduct a test of significance with the following hypotheses:
Null: = 0.20 Alternative: 0.20
the absolute tolerance is +/-0.0002
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the absolute tolerance is +/-0.0005
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A recent study examined hearing loss data for 1769 U.S. teenagers. In this sample, 331 were found to have some level of hearing loss. News of this study spread quickly, with many news articles blaming the prevalence of hearing loss on the higher use of ear buds by teens. At MSNBC.com (8/17/2010), Carla Johnson summarized the study with the headline: “1 in 5 U.S. teens has hearing loss, study says.” To investigate whether this is an appropriate or a misleading headline, you will conduct a test of significance with the following hypotheses:
Null: = 0.20 Alternative: 0.20
Explanation / Answer
Result:
A recent study examined hearing loss data for 1769 U.S. teenagers. In this sample, 331 were found to have some level of hearing loss. News of this study spread quickly, with many news articles blaming the prevalence of hearing loss on the higher use of ear buds by teens. At MSNBC.com (8/17/2010), Carla Johnson summarized the study with the headline: “1 in 5 U.S. teens has hearing loss, study says.” To investigate whether this is an appropriate or a misleading headline, you will conduct a test of significance with the following hypotheses:
Null: = 0.20 Alternative: 0.20
Use the Theory-Based Inference applet to determine a p-value. Round your answer to 4 decimal places, e.g. 0.7534.
P=0.1753
the absolute tolerance is +/-0.0002
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Based on your p-value, there is strong evidence that the proportion of all U.S. teens with some hearing loss is different than 1 in 5 (or 20%).
True
Answer: False
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Using the applet, find a 95% confidence interval for the proportion of U.S. teens that have some hearing loss. Round your answer to 4 decimal places, e.g. 0.7534.
The 95% confidence interval is 0.1689 to 0.2053 .
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What is the margin of error for your confidence interval from part (d) of this question? Round your answer to 4 decimal places, e.g. 0.7534.
Margin of error = 0.0182
the absolute tolerance is +/-0.0005
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Based on your confidence interval, 0.20 is a plausible value for the proportion of the population that has some hearing loss.
Answer: True
False
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Based on your p-value, 0.20 is a plausible value for the proportion of the population that has some hearing loss.
Answer; True
False
Z Test of Hypothesis for the Proportion
Data
Null Hypothesis p =
0.2
Level of Significance
0.05
Number of Items of Interest
331
Sample Size
1769
Intermediate Calculations
Sample Proportion
0.187111362
Standard Error
0.0095
Z Test Statistic
-1.3552
Two-Tail Test
Lower Critical Value
-1.9600
Upper Critical Value
1.9600
p-Value
0.1753
Do not reject the null hypothesis
Confidence Interval Estimate for the Proportion
Data
Sample Size
1769
Number of Successes
331
Confidence Level
95%
Intermediate Calculations
Sample Proportion
0.187111362
Z Value
-1.9600
Standard Error of the Proportion
0.0093
Interval Half Width
0.0182
Confidence Interval
Interval Lower Limit
0.1689
Interval Upper Limit
0.2053
A recent study examined hearing loss data for 1769 U.S. teenagers. In this sample, 331 were found to have some level of hearing loss. News of this study spread quickly, with many news articles blaming the prevalence of hearing loss on the higher use of ear buds by teens. At MSNBC.com (8/17/2010), Carla Johnson summarized the study with the headline: “1 in 5 U.S. teens has hearing loss, study says.” To investigate whether this is an appropriate or a misleading headline, you will conduct a test of significance with the following hypotheses:
Null: = 0.20 Alternative: 0.20