/*! 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 12 In a certain region prepackaged ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a certain region prepackaged products labeled \(500 \mathrm{~g}\) must contain on average at least 500 grams of the product, and at least \(90 \%\) of all packages must weigh at least 490 grams. In a random sample of 300 packages, 288 weighed at least 490 grams. a. Give a point estimate of the proportion of all packages that weigh at least 490 grams. b. Verify that the sample is sufficiently large to use it to construct a confidence interval for that proportion. c. Construct a \(99.8 \%\) confidence interval for the proportion of all packages that weigh at least 490 grams.

Short Answer

Expert verified
a. 0.96; b. Sample is large enough; c. (0.941, 0.979).

Step by step solution

01

Finding the Point Estimate

We have a sample of 300 packages, of which 288 weigh at least 490 grams. The point estimate for the proportion of packages that weigh at least 490 grams is calculated as \( \hat{p} = \frac{288}{300} \). This gives us \( \hat{p} = 0.96 \). This means that 96% of the sampled packages weigh at least 490 grams.
02

Checking Sample Size Sufficiency

To verify that the sample is large enough to construct a confidence interval, we ensure that \( np \geq 5 \) and \( n(1-p) \geq 5 \). Using the point estimate \( \hat{p} = 0.96 \): 1. \( n\hat{p} = 300 \times 0.96 = 288 \geq 5 \)2. \( n(1-\hat{p}) = 300 \times 0.04 = 12 \geq 5 \)Both conditions are satisfied, so the sample is sufficiently large.
03

Calculating the 99.8% Confidence Interval

For a 99.8% confidence interval, we determine the critical value using the standard normal distribution table, which gives \( z \approx 3.090 \). We use the formula:\[CI = \hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]Substituting the values:\[CI = 0.96 \pm 3.090 \sqrt{\frac{0.96 \times 0.04}{300}}\]\[= 0.96 \pm 0.019\]Thus, the confidence interval is \( (0.941, 0.979) \).

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.

Point Estimate
When we talk about a point estimate in statistics, we refer to a single value that serves as an estimate of a population parameter. In simpler terms, it is a best guess we can derive from the sample data about the characteristic of a larger group.
For the exercise in question, the point estimate is the proportion of packages weighing at least 490 grams from our sample data. Given that 288 out of 300 packages weighed as much, our point estimate
  • is calculated by dividing the favorable cases (packages meeting the weight requirement) by the total number of samples, resulting in \( \hat{p} = \frac{288}{300} \).
  • This calculation gives \( \hat{p} = 0.96 \), meaning 96% of the sampled packages meet the weight condition.
This point estimate provides us a snapshot of what we can expect in the whole population based on our sample.
Sample Size Sufficiency
Before constructing a confidence interval, it's crucial to ensure that the sample size is adequate. This is a fundamental concept in statistics because it affects the reliability and validity of our estimates.
In our example, the conditions for sample size sufficiency are:
  • First, we calculate \( n\hat{p} \) as \( 300 \times 0.96 = 288 \).
  • Next, we find \( n(1-\hat{p}) \) as \( 300 \times 0.04 = 12 \).
Both calculations ensure that the criteria \( n\hat{p} \geq 5 \) and \( n(1-\hat{p}) \geq 5 \) are satisfied, which confirms our sample is large enough.
If these conditions were not met, our confidence interval could be unreliable.
Critical Value
The critical value is a crucial component when calculating a confidence interval. It tells us the number of standard deviations we need to go from the mean to capture a desired percentage of the data in a normal distribution.
For constructing a confidence interval, especially at a high confidence level like 99.8%, the critical value must be carefully selected using a z-table or standard normal distribution table.
  • In our case, for a 99.8% confidence interval, the critical value (\( z \)) is approximately 3.090.
  • The critical value allows us to determine the margin of error in our confidence interval calculation.
We then use the formula:\[CI = \hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]Inserting the values from the exercise gives a confidence interval of \((0.941, 0.979)\), which interprets to having a high degree of certainty that the actual proportion of all packages meeting the weight requirement falls within this range. This application of critical value helps to express the reliability and precision of our estimations.

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

The number of trips to a grocery store per week was recorded for a randomly selected collection of households, with the results shown in the table. $$ \begin{array}{lllllllll} 2 & 2 & 2 & 1 & 4 & 2 & 3 & 2 & 5 & 4 \\ 2 & 3 & 5 & 0 & 3 & 2 & 3 & 1 & 4 & 3 \\ 3 & 2 & 1 & 6 & 2 & 3 & 3 & 2 & 4 & 4 \end{array} $$ Construct a \(95 \%\) confidence interval for the average number of trips to a grocery store per week of all households.

Estimate the minimum sample size needed to form a confidence interval for the mean of a population having the standard deviation shown, meeting the criteria given. a. \(\quad \sigma=30,95 \%\) confidence, \(E=10\) b. \(\quad \sigma=30,99 \%\) confidence, \(E=10\) c. \(\quad \sigma=30,95 \%\) confidence, \(E=5\)

You wish to estimate the current mean birth weight of all newborns in a certain region, to within 1 ounce (1/16 pound) and with \(95 \%\) confidence. A sample will cost \(\$ 400\) plus \(\$ 1.50\) for every newborn weighed. You believe the standard deviations of weight to be no more than 1.25 pounds. You have \(\$ 2,500\) to spend on the study. a. Can you afford the sample required? b. If not, what are your options?

A random sample of size 256 is drawn from a population whose distribution, mean, and standard deviation are all unknown. The summary statistics are \(x-=1011\) and \(s=34\). a. Construct a \(90 \%\) confidence interval for the population mean \(\mu\). b. Construct a \(99 \%\) confidence interval for the population mean \(\mu\). c. Comment on why one interval is longer than the other.

Estimate the minimum sample size needed to form a confidence interval for the proportion of a population that has a particular characteristic, meeting the criteria given. a. \(95 \%\) confidence, \(E=0.02\) b. \(99 \%\) confidence, \(E=0.02\) c. \(95 \%\) confidence, \(E=0.01\)

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.