/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Carbohydrate Grams in Commercial... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Carbohydrate Grams in Commercial Subs The num- ber of grams of carbohydrates in various commercially prepared 7 -inch subs is recorded below. The population standard deviation is 6.46 . Estimate the mean number of carbs in all similarly sized subs with \(95 \%\) confidence. $$ \begin{array}{lllllllll}{63} & {67} & {61} & {64} & {51} & {42} & {56} & {70} & {61} \\ {55} & {60} & {55} & {57} & {60} & {60} & {66} & {55} & {58} \\\ {70} & {65} & {49} & {51} & {61} & {54} & {50} & {55} & {56} \\ {53} & {65} & {68} & {63} & {48} & {54} & {56} & {57}\end{array} $$

Short Answer

Expert verified
The 95% confidence interval for the mean number of carbs is approximately (52.25, 56.47).

Step by step solution

01

Calculate the Sample Mean

First, we need to find the sample mean of the carbohydrates in the subs. Add up all the carbohydrate values and divide by the total number of subs. The values are: 63, 67, 61, 64, 51, 42, 56, 70, 61, 55, 60, 55, 57, 60, 60, 66, 55, 58, 70, 65, 49, 51, 61, 54, 50, 55, 56, 53, 65, 68, 63, 48, 54, 56, and 57. Summing these gives a total of 1957, and there are 36 subs. Calculate the mean: \( \bar{x} = \frac{1957}{36} \approx 54.36 \).
02

Determine the Z-score for 95% Confidence Level

For a 95% confidence level, the Z-score that corresponds to the two-tailed confidence interval is approximately 1.96. This value is obtained from the standard normal distribution table.
03

Calculate the Standard Error

The standard error of the mean (SE) is calculated using the formula: \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation (6.46) and \( n \) is the sample size (36). Calculate the standard error: \( SE = \frac{6.46}{\sqrt{36}} = \frac{6.46}{6} \approx 1.077 \).
04

Compute the Confidence Interval

Utilize the formula for the confidence interval: \( \bar{x} \pm Z \times SE \). Using the sample mean (54.36), Z-score (1.96), and standard error (1.077), calculate the confidence interval: \[ 54.36 \pm 1.96 \times 1.077 \approx 54.36 \pm 2.11 \]. Therefore, the confidence interval is approximately \( (52.25, 56.47) \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Population Standard Deviation
Understanding population standard deviation is crucial when estimating how much variation exists in a population. It is a measurement of the spread of a set of numbers. In our exercise, the population standard deviation of carbohydrates in subs is given as 6.46. This means that the amount of carbohydrates varies approximately 6.46 grams around the average carb content in all similar-sized subs.

Here’s why it matters:
  • It helps us understand the reliability of our data.
  • It gives us an idea of how much individual data points deviate from the mean.
  • It's essential for calculating other statistical measures like the standard error.
By knowing the population standard deviation, we can make informed estimates and predictions about our data set.
Sample Mean Calculation
Calculating the sample mean is a foundational step in many statistical analyses. In this exercise, the sample mean represents the average number of carbohydrates in our selected sample of subs. To find this, we summed all the carbohydrate values given for the 36 subs and then divided by the total number.

The calculation went as follows:
  • Total carbs = 1957 (sum of all subs' carb values)
  • Number of subs = 36
  • Sample Mean, \( \bar{x} = \frac{1957}{36} \approx 54.36 \)
The sample mean is essentially an estimate of the population mean of subs. It's a point estimate, meaning it's our best single guess for the true mean of all subs, but without additional context like a confidence interval, it's just a point in the entire landscape of possible values.
Standard Error
The standard error of the mean (SE) measures how much the sample mean (the average) of our data is expected to vary from the true population mean. The smaller the standard error, the closer our sample mean is likely to be to the true population mean.

Standard error is calculated using the formula:
  • \( SE = \frac{\sigma}{\sqrt{n}} \)
  • Where \( \sigma = 6.46 \) (population standard deviation)
  • And \( = 36 \) (sample size)
  • This gives us \( SE = \frac{6.46}{\sqrt{36}} = \frac{6.46}{6} \approx 1.077 \)
The standard error is a key component when calculating confidence intervals, as it provides a margin of error for our sample's mean estimate.
Z-score for Confidence Levels
In statistics, a Z-score is a measure that describes a value's relationship to the mean of a group of values, but it also plays a crucial role in confidence intervals. Specifically, the Z-score determines how many standard deviations a data point is from the mean. In confidence interval estimation, it helps us know how confident we can be in the results.

For a 95% confidence level, as in this exercise, the corresponding Z-score is 1.96. This score is derived from the standard normal distribution.
  • The Z-score tells us the range within which we can expect the true population parameter to fall, with a certain level of confidence.
  • It makes it possible to calculate the margin of error when combined with the standard error.
  • This concept is crucial for confirming how widely a variable can spread around a sample mean.
Using the Z-score, along with the sample mean and standard error, allows us to compute the confidence interval. The interval gives us a range of values, believed to encompass the true population mean with a specified level of confidence.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Holiday Gifts A survey of 100 Americans found that \(68 \%\) said they find it hard to buy holiday gifts that convey their true feelings. Find the \(90 \%\) confidence interval of the population proportion.

Women Representatives in State Legislature A state representative wishes to estimate the mean number of women representatives per state legislature. A random sample of 17 states is selected, and the number of women representatives is shown. Based on the sample, what is the point estimate of the mean? Find the \(90 \%\) confidence interval of the mean population. (Note: The population mean is actually \(31.72, \text { or about } 32 .) \mathrm{Com}\)pare this value to the point estimate and the confidence interval. There is something unusual about the data. Describe it and state how it would affect the confidence interval. $$ \begin{array}{ccccc}{5} & {33} & {35} & {37} & {24} \\ {31} & {16} & {45} & {19} & {13} \\ {18} & {29} & {15} & {39} & {18} \\ {58} & {132} & {}\end{array} $$

Hospital Noise Levels In the hospital study cited in Exercise \(19,\) the mean noise level in 171 randomly se- lected ward areas was 58.0 decibels, and the population standard deviation was 4.8 . Find the \(90 \%\) confidence interval of the true mean.

Thunderstorm Speeds A meteorologist who sampled 13 randomly selected thunderstorms found that the average speed at which they traveled across a certain state was 15.0 miles per hour. The standard deviation of the sample was 1.7 miles per hour. Find the \(99 \%\) confidence interval of the mean. If a meteorologist wanted to use the highest speed to predict the times it would take storms to travel across the state in order to issue warnings, what figure would she likely use?

Belief in Haunted Places A random sample of 205 college students was asked if they believed that places could be haunted, and 65 responded yes. Estimate the true proportion of college students who believe in the possibility of haunted places with \(99 \%\) confidence. According to Time magazine, \(37 \%\) of all Americans believe that places can be haunted.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.