/*! 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} Q3E Suppose that a random sample of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that a random sample of \({\rm{50}}\) bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let \({\rm{\mu }}\) denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting \({\rm{95\% }}\)confidence interval is \({\rm{(7}}{\rm{.8,9}}{\rm{.4)}}\). a. Would a \({\rm{90\% }}\) confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning. b. Consider the following statement: There is a \({\rm{95\% }}\) chance that \({\rm{\mu }}\) is between \({\rm{7}}{\rm{.8}}\) and \({\rm{9}}{\rm{.4}}\). Is this statement correct? Why or why not? c. Consider the following statement: We can be highly confident that \({\rm{95\% }}\) of all bottles of this type of cough syrup have an alcohol content that is between \({\rm{7}}{\rm{.8}}\) and \({\rm{9}}{\rm{.4}}\). Is this statement correct? Why or why not? d. Consider the following statement: If the process of selecting a sample of size \({\rm{50}}\) and then computing the corresponding \({\rm{95\% }}\) interval is repeated \({\rm{100}}\)times, \({\rm{95}}\) of the resulting intervals will include \({\rm{\mu }}\). Is this statement correct? Why or why not?

Short Answer

Expert verified

(a) The broader the interval, the higher the confidence level.

(b) The statement is incorrect.

(c) The statement is incorrect.

(d) The statement is incorrect.

Step by step solution

01

Define interval

An interval is a set of numbers that includes all the real numbers between the two endpoints of the interval.

02

Explanation


(a) When a normal population is given,

\({\rm{100(1 - \alpha )\% confidence interval}}\)

the mean is calculated using,

\(\left( {{\rm{\bar x - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}{\rm{,\bar x + }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{\sigma }}}{{\sqrt {\rm{n}} }}} \right)\)

when it is known what the value\({{\rm{\sigma }}^{\rm{2}}}\)is.

The broader the interval, the higher the confidence level.

As a result, a \({\rm{90\% }}\) confidence interval is narrower than a \({\rm{95\% }}\) confidence interval. Because \({{\rm{z}}_{{\rm{\alpha /2}}}}\) for \({\rm{90\% }}\) confidence level is smaller than \({{\rm{z}}_{{\rm{\alpha /2}}}}\) for \({\rm{95\% }}\) confidence level \({\rm{(1}}{\rm{.645 < 1}}{\rm{.96)}}\), the confidence interval will be shorter.

03

Explanation


(b) This assertion is incorrect.

In general, there is \({\rm{95\% }}\) confidence in the technique when the confidence interval is formed from a sample, whether the mean \({\rm{\mu }}\) belongs to it or not.

04

Explanation


(c) This assertion is incorrect.

The population mean \({\rm{\mu }}\), not the population values, is the subject of the confidence interval.

05

Explanation


(d) This assertion is incorrect.

It isn't required. The sample mean \({\rm{\mu }}\) is predicted to fall within \({\rm{95}}\) and \({\rm{100}}\) percent confidence intervals. In theory, if you repeated the process infinitely instead of \({\rm{100}}\) times, the sample mean \({\rm{\mu }}\) would fall into \({\rm{95\% }}\) of those confidence ranges.

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!

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

Assume that the helium porosity (in percentage) of coal samples taken from any particular seam is normally distributed with true standard deviation .75.

a. Compute a 95% CI for the true average porosity of a certain seam if the average porosity for 20 specimens from the seam was 4.85.

b. Compute a 98% CI for true average porosity of another seam based on 16 specimens with a sample average porosity of 4.56.

c. How large a sample size is necessary if the width of the 95% interval is to be .40?

d. What sample size is necessary to estimate true average porosity to within .2 with 99% confidence?

On the basis of extensive tests, the yield point of a particular type of mild steel-reinforcing bar is known to be normally distributed with\({\rm{\sigma = 100,}}\)The composition of bars has been slightly modified, but the modification is not believed to have affected either the normality or the value of\({\rm{\sigma }}\).

a. Assuming this to be the case, if a sample of 25 modified bars resulted in a sample average yield point of 8439 lb, compute a 90% CI for the true average yield point of the modified bar.

b. How would you modify the interval in part (a) to obtain a confidence level of 92%?

Determine the t critical value for a lower or an upper confidence bound for each of the situations

\(\begin{array}{*{20}{l}}{{\rm{a}}{\rm{. Confidence level = 95\% , df = 10}}}\\{{\rm{b}}{\rm{. Confidence level = 95\% , df = 15}}}\\{{\rm{c}}{\rm{. Confidence level = 99\% , df = 15}}}\\{{\rm{d}}{\rm{. Confidence level = 99\% , n = 5}}}\\{{\rm{\;e}}{\rm{. Confidence level = 98\% , df = 24}}}\\{{\rm{f}}{\rm{. Confidence level = 99\% , n = 38}}}\end{array}\)

Each of the following is a confidence interval for \({\rm{\mu = }}\) true average (i.e., population mean) resonance frequency (Hz) for all tennis rackets of a certain type: \({\rm{(114}}{\rm{.4,115}}{\rm{.6)(114}}{\rm{.1,115}}{\rm{.9)}}\) a. What is the value of the sample mean resonance frequency? b. Both intervals were calculated from the same sample data. The confidence level for one of these intervals is \({\rm{90\% }}\) and for the other is \({\rm{99\% }}\). Which of the intervals has the \({\rm{90\% }}\) confidence level, and why?

TV advertising agencies face increasing challenges in reaching audience members because viewing TV programs via digital streaming is gaining in popularity. The Harris poll reported on November 13, 2012, that 53% of 2343 American adults surveyed said they have watched digitally streamed TV programming on some type of device.

a. Calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult Americans who watched streamed programming up to that point in time.

b. What sample size would be required for the width of a 99% CI to be at most .05 irrespective of the value of p ˆ?

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.