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Scientists estimate that there are 10 times more bacterial cells in your body than your own body's cells, and new studies on bacteria in the gut indicate that your gut microbes might be influencing you more than you realize, having positive or negative effects on health, development, and possibly even personality and behavior. A recent study \(^{26}\) found that the average number of unique genes in gut bacteria, for a sample of 99 healthy European individuals, was 564 million, with a standard deviation of 122 million. Use the t-distribution to find and interpret a \(95 \%\) confidence interval for the mean number of unique genes in gut bacteria for European individuals.

Short Answer

Expert verified
The exact numerical calculations will provide a 95% confidence interval for the mean number of unique genes in gut bacteria for European individuals. This result means that we can be 95% confident that the interval contains the true mean number of unique genes in gut bacteria for all European individuals.

Step by step solution

01

Identify given values

The problem provides the following information: sample mean (\(x̄\)) = 564 million, sample standard deviation (s) = 122 million, sample size (n) = 99, and confidence level = 0.95 or 95%.
02

Determine the degrees of freedom

For a t-distribution, the degrees of freedom is typically calculated as the sample size minus 1. In this case, the degrees of freedom would be \(n - 1 = 99 - 1 = 98\).
03

Look up the t-value

Use a t-table or a calculator to find the t-value that corresponds to the 95% confidence level and 98 degrees of freedom. This t-value will be around 1.984, but its exact value might slightly vary depending on the table or calculator used.
04

Calculate the margin of error

Using the t-value, calculate the margin of error, which is the amount added and subtracted from the sample mean to create the confidence interval. The formula for the margin of error is \(E = t * s / \sqrt{n}\), where \(t\) is the t-score, \(s\) is the standard deviation, and \(n\) is the sample size. So, in this case, \(E \approx 1.984 * 122/\sqrt{99}\).
05

Calculate the confidence interval

The confidence interval is then calculated by subtracting and adding the margin of error from the sample mean. The formula used is \(x̄ \pm E\). Compute the exact numerical values and express the result in terms of millions.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values that's used to estimate a population parameter, like the mean. It's important because it gives us an idea of how confident we can be about the sample statistics reflecting the true population parameters. In this exercise, scientists are looking at gut bacteria genes to understand their influence on health. When we talk about a 95% confidence interval, it implies that if we repeated the study many times, 95% of the calculated confidence intervals will contain the true population mean. This helps researchers and statisticians say with a given probability that the parameter lies within that range. To determine this confidence interval using the t-distribution (as the population standard deviation is unknown and the sample size is below 30), we:
  • Identify the sample mean, which is the starting point for the estimation.
  • Calculate the margin of error, which defines the range added to and subtracted from the sample mean.
The actual confidence interval is formed by adding and subtracting this margin from the sample mean, telling us where the true mean might lie. This is a crucial concept in statistics, providing insight into the reliability of our estimates.
Degrees of Freedom
Degrees of freedom (often abbreviated as df) are crucial when working with the t-distribution, particularly for smaller sample sizes or when the population standard deviation isn't known. They provide the basis for calculating the correct t-value. Degrees of freedom can be understood as the number of independent values or quantities that can be assigned in a statistical calculation. In other words, it's the number of values that are free to vary in our dataset while ensuring that established constraints (like a certain sum) remain met. In our example, the degrees of freedom are calculated simply as the sample size minus one, which is 98 in this context ( -1=99-1=98). This accounts for the variability in the dataset. The correct degrees of freedom affect the t-value you should use when estimating the confidence interval. With the right degrees of freedom, you can look up or calculate the right t-value to use in your margin of error calculation, ensuring a more accurate confidence interval for the population mean.
Margin of Error
The margin of error (E) tells you how precise the sample estimate is. In simpler terms, it involves adding or subtracting a small amount in order to build a confidence interval around the sample mean. This provides us with an upper and lower boundary that most likely contains the true population mean.The margin of error in statistical analysis is determined using the t-distribution, especially when dealing with smaller samples or unknown population standard deviations. The margin of error is calculated using the formula:\[ E = t \times \frac{s}{\sqrt{n}} \]Here:
  • t is the t-score, representing the number of standard deviations from the mean required to encapsulate the desired level of confidence.
  • s is the sample standard deviation, reflecting the dataset's actual variations.
  • n is the sample size, showing the number of observations.
Getting accurate results requires accurate calculation and understanding of the margin of error, as it determines the extent of the interval around the sample mean. In practice, this provides researchers confidence about how much they can trust their data-driven conclusions. Ensuring clarity on the margin of error is vital for making informed decisions based on statistical data.

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Most popular questions from this chapter

Use a t-distribution to find a confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) using the relevant sample results from paired data. Give the best estimate for \(\mu_{1}-\) \(\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using \(d=x_{1}-x_{2}\). A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=3.7, s_{d}=\) 2.1, \(n_{d}=30\)

Restaurant Bill by Gender In the study described in Exercise 6.224 the diners were also chosen so that half the people at each table were female and half were male. Thus we can also test for a difference in mean meal cost between females \(\left(n_{f}=24, \bar{x}_{f}=44.46, s_{f}=15.48\right)\) and males \(\left(n_{m}=24, \bar{x}_{m}=43.75, s_{m}=14.81\right) .\) Show all details for doing this test.

Split the Bill? Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The data in SplitBill includes the cost of what each person ordered (in Israeli shekels) and the payment method (Individual or Split). Some summary statistics are provided in Table 6.20 and both distributions are reasonably bell-shaped. Use this information to test (at a \(5 \%\) level ) if there is evidence that the mean cost is higher when people split the bill. You may have done this test using randomizations in Exercise 4.118 on page 302 .

The dataset ICUAdmissions, introduced in Data 2.3 on page \(69,\) includes information on 200 patients admitted to an Intensive Care Unit. One of the variables, Status, indicates whether each patient lived (indicated with a 0 ) or died (indicated with a 1 ). Use technology and the dataset to construct and interpret a \(95 \%\) confidence interval for the proportion of ICU patients who live.

Do Babies Prefer Speech? Psychologists in Montreal and Toronto conducted a study to determine if babies show any preference for speech over general noise. \(^{61}\) Fifty infants between the ages of \(4-13\) months were exposed to both happy-sounding infant speech and a hummed lullaby by the same woman. Interest in each sound was measured by the amount of time the baby looked at the woman while she made noise. The mean difference in looking time was 27.79 more seconds when she was speaking, with a standard deviation of 63.18 seconds. Perform the appropriate test to determine if this is sufficient evidence to conclude that babies prefer actual speaking to humming.

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