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Many cardiac patients wear implanted pacemakers to control their heartbeat. A plastic connector module mounts on the top of the pacemaker. Assuming a standard deviation of 0.0015 and an approximate normal distribution, find a \(95 \%\) confidence interval for the mean of all connector modules made by a certain manufacturing company. A random sample of 75 modules has an average of 0.310 inch.

Short Answer

Expert verified
The 95% confidence interval for the mean of all connector modules is (0.30966, 0.31034)

Step by step solution

01

Identify given values

The sample size \(n\) is 75, the sample mean \(\bar{x}\) is 0.310, the population standard deviation \(\sigma\) is 0.0015, and the confidence level is 95%.
02

Determine the Z-score

Find the corresponding Z-score for a 95% confidence level. From the Z-table or by using statistical software, the Z-score that corresponds to a 95% confidence interval is approximately 1.96.
03

Use the formula for the confidence interval

The formula for a confidence interval is \(\bar{x} \pm Z * \frac{\sigma}{\sqrt{n}}\). Substitute the given values into the formula: 0.310 ± 1.96 * 0.0015/√75.
04

Calculate the margin of error

The margin of error is calculated as 1.96 * 0.0015/√75 = 0.00034.
05

Calculate the confidence interval

Finally, add and subtract the margin of error from the sample mean to get the 95% confidence interval. This results in 0.310 + 0.00034 = 0.31034 and 0.310 - 0.00034 = 0.30966. So, the 95% confidence interval for the mean of all connector modules is (0.30966, 0.31034)

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Key Concepts

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

Standard Deviation
When considering the reliability and consistency of a set of data, one of the most fundamental concepts is the standard deviation. It's a measure of how much the numbers in a data set deviate or diverge from the mean, or average, value.

For instance, if we are assessing the sizes of plastic connector modules, a small standard deviation indicates that most modules are very close to the average size. In contrast, a large standard deviation would suggest that there's a wide range of sizes — some far smaller or larger than the average.

To calculate the standard deviation, usually represented by the symbol \( \sigma \) for a population or \( s \) for a sample, you first find the difference between each data point and the mean, square those differences, then average those squared differences, and finally, take the square root of that average. In the example provided, the standard deviation of the connector modules is given as 0.0015 inches, indicating that the sizes of the modules are fairly consistent around the mean value.
Normal Distribution
The normal distribution, also known as the bell curve or Gaussian distribution, is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

In the case of the connector modules, we're told that the sizes approximately follow a normal distribution. This means we expect most of the modules to have sizes close to the mean, with fewer and fewer modules being extremely small or large. In a perfectly normal distribution, about 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% within three.

The importance of normal distribution comes into play when we make inferences about a population from a sample. As we predict what the true average size of all connector modules might be, we rely on the assumption that their sizes distribute normally to use certain statistical tools, like the Z-score and confidence interval.
Z-score
The Z-score is a statistical measure that tells us how many standard deviations an element is from the mean. It is a way of standardizing scores on a single, common scale. When we know that a distribution is approximately normal, we can use the Z-score to find out how unusual or typical a certain value is.

In your textbook example, to establish a 95% confidence interval for the mean size of the connector modules, we seek the Z-score that corresponds to the middle 95% of the normal distribution. This Z-score gives us a critical value to calculate how far from our sample mean we'd expect the true population mean to fall, with a given level of confidence. The Z-score for a 95% confidence level is commonly 1.96. This would indicate that the true mean size of all connector modules is likely to be within 1.96 standard deviations from our sample mean of 0.310 inches.

Applying this Z-score to our standard deviation and sample size, as detailed in the steps provided, helps us create a range or interval that we believe contains the true mean size with 95% certainty.

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Most popular questions from this chapter

A consumer group is interested in comparing operating costs for two different types of automobile engines. The group is able to find 15 owners whose cars have engine type \(A\) and 15 who have engine type B. All 30 owners bought their cars at roughly the same time and all have kept good records for a certain 12 month period. In addition, owners were found that drove roughly the same mileage. The cost statistics are \(y_{A}=\$ 87.00 / 1,000\) miles, \(y_{B}=\$ 75.00 / 1,000\) miles, \(s_{A}=\$ 5.99,\) and \(S B-\$ 4.85 .\) Compute a \(95 \%\) confidence interval to estimate \(\mu_{A}-\mu_{B_{1}}\) the difference in the mean operating costs. Assume normality and equal variance.

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According to USA Today (March 17. \(\lfloor 997\) ), women made up \(33.7 \%\) of the editorial staff at local TV stations in 1990 and \(36.2 \%\) in \(1994 .\) Assume 20 new employees were: hired as editorial staff. (a) Estimate the number that would have been women in each year, respectively, (b) Compute a \(95 \%\) confidence interval to see if there is evidence that the proportion of women hired as editorial staff in 1994 was higher than the proportion hired in 1990 .

An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours, find a \(96 \%\) confidence interval for the population mean of all bulbs produced by this firm.

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