The following sample of 16 measurements was selected from a population that is approximately normally distributed:

  1. Construct an 80% confidence interval for the population mean.
  2. Construct a 95% confidence interval for the population mean and compare the width of this interval with that of part a.
  3. Carefully interpret each of the confidence intervals and explain why the 80% confidence interval is narrower.

Short Answer

Expert verified

A confidence interval is described as the set of numbers observed in our collection for which we anticipate discovering the figure that best represents the whole population.

Step by step solution

01

(a) The data is given below

The calculation is given below:

n = 16X¯=97.9375s = 12.6463

df = 16-1= 15

α=0.20

The critical worth of localid="1652095550050" t=±1.341

The confidence interval is 80%:

localid="1652095596149" Xt×sn=97.9375±1.341×12.646316=97.9375±4.238=93.699,102.176

The80%confidential level93.699,102.176

02

(b) The data is given below

The calculation is given below:

n = 16X¯=97.9375s = 12.6463n = 16X¯=97.9375s = 12.6463

df = 16-1= 15

α=0.20

The critical worth of t=±1.341

The confidence interval is 80%:

Xt×sn=97.9375±2.131×12.646316=97.9375±6.739=91.199,104.676

The95%confidential level91.199,104.676

The confidence interval 80% is (93,699,102.176) is narrower than confidence interval 95% is (91.199, 104.676)

03

(c) The data is given below

As the level of confidence goes from 95% to 80%, the width of the confidence interval narrows. Since lowering the confidence threshold from 95% to 80% reduces the error margin, resulting in a smaller confidence interval. As the level of confidence lowers, so does our estimation of the genuine unknown population parameter to be around a specific interval.

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 Vaia!

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

Improving the productivity of chickens. Farmers have discovered that the more domestic chickens peck at objects placed in their environment, the healthier and more productive the chickens seem to be. White string has been found to be a particularly attractive pecking stimulus. In one experiment, 72 chickens were exposed to a string stimulus. Instead of white string, blue-colored string was used. The number of pecks each chicken took at the blue string over a specified time interval was recorded. Summary statistics for the 72 chickens were x = 1.13 pecks, s = 2.21 pecks (Applied Animal Behavior Science, October 2000).

a. Estimate population mean number of pecks made by chickens pecking at blue string using a 99% confidence interval. Interpret the result.

b. Previous research has shown that = 7.5 pecks if chickens are exposed to white string. Based on the results, part a, is there evidence that chickens are more apt to peck at white string than blue string? Explain.

Question: Explain the differences in the sampling distributions of xfor large and small samples under the following assumptions.

a. The variable of interest, x, is normally distributed.

b. Nothing is known about the distribution of the variable x.

Jitter in a water power system. Jitter is a term used to describe the variation in conduction time of a water power system. Low throughput jitter is critical to successful waterline technology. An investigation of throughput jitter in the opening switch of a prototype system (Journal of Applied Physics) yielded the following descriptive statistics on conduction time for n = 18 trials:x=334.8 nanoseconds, s = 6.3 nanoseconds. (Conduction time is defined as the length of time required for the downstream current to equal 10% of the upstream current.)

a. Construct a 95% confidence interval for the true standard deviation of conduction times of the prototype system.

b. Practically interpret the confidence interval, part a.

c. A system is considered to have low throughput jitter if the true conduction time standard deviation is less than 7 nanoseconds. Does the prototype system satisfy this requirement? Explain.

Use Table III, Appendix D to determine thet0 values foreach of the following probability statements and their respectivedegrees of freedom (df ).

a.Ptt0=.25withdf=15

b.Ptt0=.1withdf=8

c.P-t0tt0=.01withdf=19

d.P-t0tt0=.05withdf=24

Question: Furniture brand familiarity. A brand name that consumers recognize is a highly valued commodity in any industry. To assess brand familiarity in the furniture industry, NPD (a market research firm) surveyed 1,333 women who head U.S. households that have incomes of $25,000 or more. The sample was drawn from a database of 25,000 households that match the criteria listed above. Of the 10 furniture brands evaluated, La-Z-Boy was the most recognized brand; 70.8% of the respondents indicated they were “very familiar” with La-Z-Boy.

a. Describe the population being investigated by NPD.

b. In constructing a confidence interval to estimate the proportion of households that are very familiar with the La-Z-Boy brand, is it necessary to use the finite population correction factor? Explain.

c. What estimate of the standard error ofp^ should be used in constructing the confidence interval of part b?

d. Construct a 90% confidence interval for the true proportion and interpret it in the context of the problem.

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.

Sign-up for free