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In Exercises 3.51 to 3.56 , information about a sample is given. Assuming that the sampling distribution is symmetric and bell-shaped, use the information to give a \(95 \%\) confidence interval, and indicate the parameter being estimated. $$ \hat{p}=0.32 \text { and the standard error is } 0.04 \text { . } $$

Short Answer

Expert verified
The 95% confidence interval for this problem is \(0.32 - 1.96(0.04)\) to \(0.32 + 1.96(0.04)\), which simplifies to approximately [0.24, 0.40]. This interval is estimating the true population proportion \(p\), given our sample statistic \(\hat{p}\), with a 95% level of confidence.

Step by step solution

01

Identify the Sample Statistic and Standard Error

The provided sample statistic \(\hat{p}\) equals 0.32 and the standard error is 0.04.
02

Use 1.96 as the Z-Score for a 95% Confidence Interval

The Z-Score value to use for a 95% confidence interval in a normal distribution is 1.96. This is a general rule of thumb that applies because the distribution is symmetric and bell-shaped.
03

Calculate the Lower and Upper Bounds of the Confidence Interval

The formula for a 95% confidence interval is \(\hat{p}\) ± 1.96 times the standard error. Calculate the lower bound of the confidence interval by subtracting 1.96 multiplied by the standard error from the sample statistic: \(0.32 - 1.96(0.04)\). Calculate the upper bound of the confidence interval by adding 1.96 multiplied by the standard error to the sample statistic: \(0.32 + 1.96(0.04)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sampling Distribution
When dealing with statistics, the concept of sampling distribution is foundational. It refers to the probability distribution of a given statistic based on a random sample. In simpler terms, imagine repeating an experiment over and over again and recording the result each time; the pattern of these numerous results is what statisticians call a sampling distribution. It's crucial because it helps us understand how the sample relates to the population from which it was drawn.

For instance, if we were to measure the mean height of a randomly selected group of ten people (our sample) from a town (our population), and then do this repeatedly with different groups of ten, we would end up with many different sample means. These means, when plotted on a graph, would typically form a bell-shaped curve if the sample size is sufficiently large. This curve is also known as the 'normal distribution' when the sample statistic follows a normal pattern. What the sampling distribution tells us is how mean heights (or other measurements) are distributed and allows us to make inferences about the entire town's height based on our samples.
Standard Error
The standard error is a statistical term that measures the precision of an estimate from a sample. Specifically, it tells you how much the sample statistic (like the mean or proportion) is expected to fluctuate if you were to take different samples from the same population. It's essentially a way to quantify the 'margin of error' in estimation.

The standard error is calculated based on the standard deviation of the sampling distribution and the size of the sample. The formula for the standard error of the mean (SEM) is the standard deviation divided by the square root of the sample size. For proportions, as in the exercise provided, it involves the sample proportion and the size of the sample, which is commonly given as SE = √(p(1-p)/n), with p being the sample proportion and n being the sample size.

As the sample size increases, the standard error decreases. This indicates that larger samples are likely to provide more accurate estimates of the population parameter. In other words, the larger the sample size, the tighter and more precise the confidence interval becomes, signaling greater confidence in the point estimate being close to the true population parameter.
Z-Score
In relation to a confidence interval, a Z-Score is a measure of how many standard deviations an element is from the mean. When constructing a confidence interval, the Z-Score helps determine how 'far out' from the sample statistic we need to go to capture the center percentage of the sampling distribution (for example, 95% in the middle for a 95% confidence interval).

A Z-Score of 1.96 corresponds to the 97.5th percentile of a standard normal distribution, meaning that approximately 97.5% of the data falls below this value. Since a normal distribution is symmetrical, 2.5% also falls above the corresponding positive Z-Score. Together, this amounts to 95% of the data falling between the negative and positive Z-Scores.

This is why the Z-Score of 1.96 is so widely used for a 95% confidence interval in normally distributed data. It's the multiplier that, when applied to the standard error, gives the margin of error for the estimate, thereby creating the interval within which we're confident the true population parameter lies. In the step-by-step exercise provided, multiplying the standard error by the Z-Score of 1.96 on either side of the sample statistic gives us a range that we can be 95% sure contains the real population proportion.

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Most popular questions from this chapter

Effect of Overeating for One Month: Average Long-Term Weight Gain Overeating for just four weeks can increase fat mass and weight over two years later, a Swedish study shows \(^{35}\) Researchers recruited 18 healthy and normal-weight people with an average age of \(26 .\) For a four-week period, participants increased calorie intake by \(70 \%\) (mostly by eating fast food) and limited daily activity to a maximum of 5000 steps per day (considered sedentary). Not surprisingly, weight and body fat of the participants went up significantly during the study and then decreased after the study ended. Participants are believed to have returned to the diet and lifestyle they had before the experiment. However, two and a half years after the experiment, the mean weight gain for participants was 6.8 lbs with a standard error of 1.2 lbs. A control group that did not binge had no change in weight. (a) What is the relevant parameter? (b) How could we find the actual exact value of the parameter? (c) Give a \(95 \%\) confidence interval for the parameter and interpret it. (d) Give the margin of error and interpret it.

What Proportion of Adults and Teens Text Message? A study of \(n=2252\) adults age 18 or older found that \(72 \%\) of the cell phone users send and receive text messages. \({ }^{15}\) A study of \(n=800\) teens age 12 to 17 found that \(87 \%\) of the teen cell phone users send and receive text messages. What is the best estimate for the difference in the proportion of cell phone users who use text messages, between adults (defined as 18 and over) and teens? Give notation (as a difference with a minus sign) for the quantity we are trying to estimate, notation for the quantity that gives the best estimate, and the value of the best estimate. Be sure to clearly define any parameters in the context of this situation.

Give information about the proportion of a sample that agrees with a certain statement. Use StatKey or other technology to estimate the standard error from a bootstrap distribution generated from the sample. Then use the standard error to give a \(95 \%\) confidence interval for the proportion of the population to agree with the statement. StatKey tip: Use "CI for Single Proportion" and then "Edit Data" to enter the sample information. In a random sample of 400 people, 112 agree and 288 disagree.

Playing Video Games A new study provides some evidence that playing action video games strengthens a person's ability to translate sensory information quickly into accurate decisions. Researchers had 23 male volunteers with an average age of 20 look at moving arrays on a computer screen and indicate the direction in which the dots were moving \(^{33}\) Half of the volunteers ( 11 men) reported playing action video games at least five times a week for the previous year, while the other 12 reported no video game playing in the previous year. The response time and the accuracy score were both measured. A \(95 \%\) confidence interval for the mean response time for game players minus the mean response time for non-players is -1.8 to -1.2 seconds, while a \(95 \%\) confidence interval for mean accuracy score for game players minus mean accuracy score for non-players is -4.2 to +5.8 . (a) Interpret the meaning of the \(95 \%\) confidence interval for difference in mean response time. (b) Is it plausible that game players and non-game players are basically the same in response time? Why or why not? If not, which group is faster (with a smaller response time)? (c) Interpret the meaning of the \(95 \%\) confidence interval for difference in mean accuracy score. (d) Is it plausible that game players and non-game players are basically the same in accuracy? Why or whynot? If not, which group is more accurate?

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