Question

Surveys have become an integral part of our lives. Because it is so important that every citizen have the ability to interpret survey results, surveys are the focus of this project. The Gallup Poll recently conducted a survey of 2,558 U.S. adults and found that 46% of those surveyed have dealt with substance abuse in their families. 1. Use the survey results to construct a 95% confidence interval to estimate the percentage of all U.S. adults who have dealt with substance abuse. 2. Identify the margin of error for the survey. 3. Explain why it would or would not be okay for a newspaper to make the statement: "based on results from a recent survey, more than 4 out of 10 adults have dealt with substance abuse." 4. Assume that you are a newspaper reporter. Write a description of the survey results for your newspaper. 5. A common criticism of surveys is that they poll only a very small percentage of the population and therefore cannot be accurate. Is a sample of only 2,558 adults taken from a population of all adults a sample size that is too small? Write a brief explanation of why the sample size of 2,558 is or is not small.

Answer 1

Sample size = n = 2558

Sample proportion of US adults who had dealt with substance abuse in their families = p = 46% = 0.46

1) 95% confidence interval for the population proportion of adults in US who had dealt with the substance abuse in their families (P) is given by -

p*(1-P) Ip 20.05/2 n

Where, is the critical value of z for two tailed test at 0.05 level of significance = 1.96 (can be obtained from the z table by finding the z corresponding to area close to 0.05/2)

<0.05/2

Hence, the confidence interval is given by -

$[0.46 \pm 1.96*\sqrt{\frac{0.46*(1-0.46)}{2558}}]$

= $[0.46\pm 0.02]$

= [0.44, 0.48]

Hence, we are 95% confident that the population proportion of US adults who had dealt with substance abuse in their families will lie between 0.44 and 0.48

2) Margin of errror = E

=

p* (1 1-p) n

= 0.02

3) Null hypothesis (Ho) : The proportion of US adults who had dealt with substance abuse in their families is 0.4

P = 4/10 = 0.4

Alternative hypothesis (H1) : The proportion of US adults who had dealt with substance abuse in their families is more than 0.5

P > 0.4

We have 95% confidence interval for P as - [0.44, 0.48]

As the confidence interval does not include the null value of population proportion (that is 0.4), we may reject the null hypothesis.

Hence, we have sufficient evidence that more than 4 out of 10 US adults, had dealt with substance abuse in their families.

So, it would be okay for newspaper to make the statement

4) The Gallop Poll survey based on 2558 US adults had found that more than 4 out of 10 US adults had dealt with substance abuse in their families.

5) The minimum sample size required is given by:

$n = (\frac{z_{0.05/2}}{E})^{2} * p_0 * (1-p_0)$

po is the specified value of population proportion under the null hypothesis = 0.4

$= (\frac{1.96}{0.02})^{2} * 0.4 * (1-0.4)$

= 230.5

So, the minimum sample size required will be 231.

Since, the sample size taken in survey (2558) is more than 231, the sample size taken is not too small.