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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

SKILL BUILDER 1 In Exercises 3.41 to \(3.44,\) data from a sample is being used to estimate something about a population. In each case: (a) Give notation for the quantity that is being estimated. (b) Give notation for the quantity that gives the best estimate. Random samples of people in Canada and people in Sweden are used to estimate the difference between the two countries in the proportion of people who have seen a hockey game (at any level) in the past year.

Exercises 3.71 to 3.73 consider the question (using fish) of whether uncommitted members of a group make it more democratic. It has been argued that individuals with weak preferences are particularly vulnerable to a vocal opinionated minority. However, recent studies, including computer simulations, observational studies with humans, and experiments with fish, all suggest that adding uncommitted members to a group might make for more democratic decisions by taking control away from an opinionated minority. \({ }^{36}\) In the experiment with fish, golden shiners (small freshwater fish who have a very strong tendency to stick together in schools) were trained to swim toward either yellow or blue marks to receive a treat. Those swimming toward the yellow mark were trained more to develop stronger preferences and became the fish version of individuals with strong opinions. When a minority of five opinionated fish (wanting to aim for the yellow mark) were mixed with a majority of six less opinionated fish (wanting to aim for the blue mark), the group swam toward the minority yellow mark almost all the time. When some untrained fish with no prior preferences were added, however, the majority opinion prevailed most of the time. \({ }^{37}\) Exercises 3.71 to 3.73 elaborate on this study. How Often Does the Fish Majority Win? In a school of fish with a minority of strongly opinionated fish wanting to aim for the yellow mark and a majority of less passionate fish wanting to aim for the blue mark, as described under Fish Democracies above, a \(95 \%\) confidence interval for the proportion of times the majority wins (they go to the blue mark) is 0.09 to \(0.26 .\) Interpret this confidence interval. Is it plausible that fish in this situation are equally likely to go for either of the two options?

Topical Painkiller Ointment The use of topical painkiller ointment or gel rather than pills for pain relief was approved just within the last few years in the US for prescription use only. \({ }^{13}\) Insurance records show that the average copayment for a month's supply of topical painkiller ointment for regular users is \$30. A sample of \(n=75\) regular users found a sample mean copayment of \(\$ 27.90\). (a) Identify each of 30 and 27.90 as a parameter or a statistic and give the appropriate notation for each. (b) If we take 1000 samples of size \(n=75\) from the population of all copayments for a month's supply of topical painkiller ointment for regular users and plot the sample means on a dotplot, describe the shape you would expect to see in the plot and where it would be centered. (c) How many dots will be on the dotplot you described in part (b)? What will each dot represent?

SKILL BUILDER 1 In Exercises 3.41 to \(3.44,\) data from a sample is being used to estimate something about a population. In each case: (a) Give notation for the quantity that is being estimated. (b) Give notation for the quantity that gives the best estimate. A random sample of maple trees in a forest is used to estimate the mean base circumference of all maple trees in the forest.

In estimating the mean score on a fitness exam, we use an original sample of size \(n=30\) and a bootstrap distribution containing 5000 bootstrap samples to obtain a \(95 \%\) confidence interval of 67 to \(73 .\) In Exercises 3.106 to 3.111 , a change in this process is described. If all else stays the same, which of the following confidence intervals \((A, B,\) or \(C)\) is the most likely result after the change: \(\begin{array}{ll}A .66 \text { to } 74 & B .67 \text { to } 73\end{array}\) C. 67.5 to 72.5 Using 1000 bootstrap samples for the distribution.

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