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Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was \(5.2 .\) Provide \(90 \%\) and \(95 \%\) confidence intervals for the population mean number of weekly customer contacts for the sales personnel.

Short Answer

Expert verified
The 90% confidence interval for the population mean number of weekly customer contacts for sales personnel is approximately (18.4216, 20.5784), and the 95% confidence interval is approximately (18.2085, 20.7915).

Step by step solution

01

Identify given values

In this exercise, we are given the following values: - Sample size (n) = 65 - Sample mean (xÌ…) = 19.5 - Sample standard deviation (s) = 5.2
02

Calculate the standard error

The standard error (SE) is the standard deviation divided by the square root of the sample size. We will use this formula to calculate the standard error: SE = s / sqrt(n) Plugging in the given values: SE = 5.2 / sqrt(65) ≈ 0.6464
03

Find the critical values for 90% and 95% confidence intervals

To find the critical values, we can use t-distribution since the population standard deviation is unknown. We will use the degrees of freedom (df), which are equal to the sample size minus 1: df = n - 1 = 65 - 1 = 64 For a 90% confidence level, the t-score (t90) is found using a t-distribution table or an online calculator. We find that: t90 ≈ 1.667 Similarly, for a 95% confidence level, the t-score (t95) is: t95 ≈ 1.998
04

Calculate the margin of error for 90% and 95% confidence intervals

We will use the formula to calculate the margin of error for both confidence intervals: Margin of error = t * SE For 90% confidence interval: Margin of error90 = t90 * SE = 1.667 * 0.6464 ≈ 1.0784 For 95% confidence interval: Margin of error95 = t95 * SE = 1.998 * 0.6464 ≈ 1.2915
05

Calculate the confidence intervals

Finally, we will calculate the confidence intervals using the sample mean and the margin of error: Confidence interval = (xÌ… - margin of error, xÌ… + margin of error) For 90% confidence interval: Confidence interval90 = (19.5 - 1.0784, 19.5 + 1.0784) = (18.4216, 20.5784) For 95% confidence interval: Confidence interval95 = (19.5 - 1.2915, 19.5 + 1.2915) = (18.2085, 20.7915) In conclusion, we can be 90% confident that the true population mean number of weekly customer contacts for sales personnel lies between 18.4216 and 20.5784, and 95% confident that it lies between 18.2085 and 20.7915.

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Key Concepts

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

Sample Mean
The sample mean is a crucial component in statistical analysis, especially when trying to estimate the average behavior of an entire population. In the provided exercise, the sample mean is given as 19.5. This value represents the average number of customer contacts made by the sample of sales personnel over a week.

The sample mean is calculated by summing all the observed values and dividing by the total number of observations. It provides a single measure that is easy to work with and helps in computing confidence intervals.

Using the sample mean, we can estimate the population mean, which is what we assume to be true for the whole group (in this case, all sales personnel). The sample mean acts as the anchor for computing the confidence interval, helping us determine the range in which the population mean likely lies.
Standard Error
The standard error (SE) is an essential statistic that quantifies the variability of the sample mean from the population mean. In other words, it helps us understand how much the sample mean might fluctuate from the actual mean if we were to take different samples.

In the exercise, the standard error was calculated using the formula:
  • SE = \( \frac{s}{\sqrt{n}} \)
where \( s \) is the sample standard deviation (5.2), and \( n \) is the sample size (65). By plugging in these values, we get an SE of approximately 0.6464.

The SE is a critical component when constructing confidence intervals, as it directly impacts the width of the interval. A smaller standard error indicates more reliable and precise sample mean estimates.
Margin of Error
The margin of error provides a range around the sample mean within which we can be confident that the true population mean lies. It takes into account the variability captured by the standard error and the desired level of confidence.

To calculate the margin of error, we multiply the standard error (SE) by the critical value (t-score) obtained from the t-distribution. For the 90% confidence level, the margin of error is calculated as \( 1.667 \times 0.6464 \approx 1.0784 \), and for the 95% confidence level, it is \( 1.998 \times 0.6464 \approx 1.2915 \).

This measure aids in creating the confidence interval and communicates the uncertainty in our estimation. The larger the margin of error, the wider the confidence interval and the more uncertainty we have about where the true population mean might lie.
t-distribution
The t-distribution is pivotal in statistical analyses involving small sample sizes or when the population standard deviation is unknown. It allows us to derive more accurate critical values for constructing confidence intervals.

In our exercise, the t-distribution is used because the population standard deviation is not provided. The t-distribution is similar to the normal distribution but has heavier tails, meaning it deals with the uncertainty of having a small sample size.

We determine the t-score based on the desired confidence level and the degrees of freedom (df), which is the sample size minus one (\( n - 1 = 64 \)). Using this, we found the critical t-values as approximately 1.667 for 90% confidence and 1.998 for 95% confidence.

By utilizing the t-distribution, we ensure that we account for the extra variability and provide more accurate confidence intervals.

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Most popular questions from this chapter

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