/*! 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 29 Average temperature in Florida A... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Average temperature in Florida According to the National Centers for Environmental Information, the mean March monthly average temperature in Florida, for the years 1895 to 2016 (a sample of 122 observations), is \(64.264^{\circ} \mathrm{F}\) with a standard deviation of \(3.109^{\circ} \mathrm{F}\) (Source: www.ncdc.noaa.gov). a. What is the point estimate of the monthly average temperature for March in Florida? b. Find the standard error of the sample mean. c. The \(95 \%\) confidence interval is (63.707,64.821) . Interpret it. d. Is it plausible that the population mean of the monthly average temperature for March in Florida could be estimated to \(\mu=63^{\circ} \mathrm{F} ?\) Explain.

Short Answer

Expert verified
The point estimate is 64.264°F. The standard error is approximately 0.281. A mean of 63°F is not plausible as it falls outside the confidence interval.

Step by step solution

01

Identify the Point Estimate

The point estimate of the monthly average temperature for March in Florida is simply the sample mean. According to the data, the sample mean is \(64.264^{\circ} \mathrm{F}\).
02

Calculate the Standard Error

The standard error of the sample mean is calculated using the formula \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = 3.109^{\circ} \mathrm{F} \) is the standard deviation and \( n = 122 \) is the sample size. Thus, \( SE = \frac{3.109}{\sqrt{122}} \approx 0.281\).
03

Interpret the Confidence Interval

The 95% confidence interval (63.707, 64.821) implies that we are 95% confident that the true population mean of the monthly average temperature for March in Florida falls within this interval.
04

Evaluate Plausibility of Given Population Mean

To determine if \( \mu = 63^{\circ} \mathrm{F} \) is a plausible estimate, we check if this value is within the confidence interval. Since \( 63^{\circ} \mathrm{F} \) is not in the interval (63.707, 64.821), it is unlikely to be the true population mean.

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
The point estimate is a single value that serves as a best guess for a statistical parameter, such as the mean of a population. When dealing with data, we often use the sample mean as a point estimate of the population mean. This is because the sample mean is calculated directly from the sample data and gives us an idea of what the average of the entire population might be. In the exercise about Florida's March temperatures, the point estimate of the monthly average temperature is simply the sample mean, which is calculated as the sum of all observations divided by the number of observations. According to the data given, this point estimate is \(64.264^{\circ} \mathrm{F}\). This represents our best estimate of what the true average March temperature in Florida is likely to be.
Standard Error
The standard error (SE) is a measure of the amount of variability in a statistic, such as the sample mean, due to the random nature of sampling. It provides an indication of how much the sample mean might differ from the true population mean, purely by chance. The standard error is particularly useful because it helps quantify the accuracy of a point estimate. Calculating the standard error involves dividing the standard deviation of the sample by the square root of the sample size \(n\): \[ SE = \frac{\sigma}{\sqrt{n}} \]Given a standard deviation \(\sigma = 3.109^{\circ} \mathrm{F}\) and a sample size \(n = 122\), we compute the standard error of the sample mean as:
  • \(SE = \frac{3.109}{\sqrt{122}} \approx 0.281\).
This means there is an approximate deviation of \(0.281^{\circ} \mathrm{F}\) from the true population mean due to sampling variability.
Confidence Interval
A confidence interval (CI) gives a range within which we expect the true population parameter, such as the population mean, to lie. It is calculated using the point estimate and the standard error, and it reflects the level of certainty we have about our estimate. A common confidence level used is 95%, meaning we are 95% confident that the true mean lies within this range.Given a point estimate \(64.264^{\circ} \mathrm{F}\) and a standard error of \(0.281\), the observed 95% confidence interval for Florida's March average temperature is \((63.707, 64.821)\). This suggests that, based on our sample, there is a 95% probability that the true population mean of March temperatures in Florida falls within this range.
  • This range is critical as it directly informs us about the reliability and precision of our point estimate.
Population Mean
The population mean \(\mu\) is the true average of a particular characteristic across the entire population. However, since it's rarely feasible to collect data from an entire population, we rely on sample data to estimate it.One of the exercises' questions was to check if an alternative population mean, such as \(63^{\circ} \mathrm{F}\), is plausible given the data. To determine this, we see if this hypothetical mean falls within our calculated confidence interval.
  • In our case, \(63^{\circ} \mathrm{F}\) does not lie within the interval \((63.707, 64.821)\).
  • This leads us to conclude it's unlikely that the true population mean is \(63^{\circ} \mathrm{F}\).
Assessing whether or not a hypothesized population mean falls within the confidence interval is a standard practice for evaluating its plausibility.

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

Abstainers The Harvard study mentioned in the previous exercise estimated that \(19 \%\) of college students abstain from drinking alcohol. To estimate this proportion in your school, how large a random sample would you need to estimate it to within 0.05 with probability \(0.95,\) if before conducting the study a. You are unwilling to guess the proportion value at your school? b. You use the Harvard study as a guideline? c. Use the results from parts a and b to explain why strategy (a) is inefficient if you are quite sure you'll get a sample proportion that is far from 0.50 .

The Google Play app store for smartphones offers hundreds of games to download for free or for a small fee. The ones for which a fee is charged are called paid games. For a random sample of five paid games taken in July 2014 on the Google platform, the following fees were charged: \(\$ 1.09, \$ 4.99, \$ 1.99, \$ 1.99, \$ 2.99 .\) a. Find a point estimate of the mean fee for paid games available on Google's platform. b. The margin of error at the \(95 \%\) confidence level for this point estimate is \(\$ 1.85 .\) Explain what this means.

A national television network takes an exit poll of 1400 voters after each has cast a vote in a state gubernatorial election. Of them, 660 say they voted for the Democratic candidate and 740 say they voted for the Republican candidate. a. Treating the sample as a random sample from the population of all voters, would you predict the winner? Base your decision on a \(95 \%\) confidence interval. b. Base your decision on a \(99 \%\) confidence interval. Explain why you need stronger evidence to make a prediction when you want greater confidence.

A national survey was conducted by the Pew Research Center (www.people-press. org) between February \(18-21,2016 .\) Among 1002 participating adults, \(51 \%\) said that Apple Inc. should assist the FBI in their investigations by unlocking the iPhone used by one of the suspects in the San Bernardino terrorist attacks. Based on these data, can we conclude that more than half of Americans support the Department of Justice over Apple Inc. in this dispute over unlocking the concerned iPhone? Explain.

Birth weights of elephants The birth weights (in kilograms) of five elephants, selected randomly, are 133,120 , 97,106,124 (Source: www.elephant.se). a. Using the web app, software or a calculator, verify that the \(95 \%\) confidence interval for the population mean is (98.11,133.89) b. Name two things you could do to get a narrower interval than the one in part a. c. Construct a \(99 \%\) confidence interval. Why is it wider than the \(95 \%\) interval? d. On what assumptions is the interval in part a based? Explain how important each assumption is.

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.