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Chapter 7: Q3E (page 284)

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.

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Most popular questions from this chapter

Example 1.11 introduced the accompanying observations on bond strength.

11.5 12.1 9.9 9.3 7.8 6.2 6.6 7.0

13.4 17.1 9.3 5.6 5.7 5.4 5.2 5.1

4.9 10.7 15.2 8.5 4.2 4.0 3.9 3.8

3.6 3.4 20.6 25.5 13.8 12.6 13.1 8.9

8.2 10.7 14.2 7.6 5.2 5.5 5.1 5.0

5.2 4.8 4.1 3.8 3.7 3.6 3.6 3.6

a.Estimate true average bond strength in a way that conveys information about precision and reliability.

(Hint: \(\sum {{{\bf{x}}_{\bf{i}}}} {\bf{ = 387}}{\bf{.8}}\) and \(\sum {{{\bf{x}}^{\bf{2}}}_{\bf{i}}} {\bf{ = 4247}}{\bf{.08}}\).)

b. Calculate a 95% CI for the proportion of all such bonds whose strength values would exceed 10.

Let \({\mathbf{0}} \leqslant {\mathbf{\gamma }} \leqslant {\mathbf{\alpha }}\). Then a 100(1 - 伪)% CI for 碌 when n is large is

\(\left( {\overline {\bf{x}} {\bf{ - }}{{\bf{z}}_{\bf{\gamma }}}{\bf{.}}\frac{{\bf{s}}}{{\sqrt {\bf{n}} }}{\bf{,}}\overline {\bf{x}} {\bf{ + }}{{\bf{z}}_{{\bf{\alpha - \gamma }}}}{\bf{ \times }}\frac{{\bf{s}}}{{\sqrt {\bf{n}} }}} \right)\)

The choice 蠏 =伪/2 yields the usual interval derived in Section 7.2; if 蠏 鈮� 伪/2, this interval is not symmetric about \(\overline x \). The width of this interval is \({\mathbf{w = s(}}{{\mathbf{z}}_{\mathbf{\gamma }}}{\mathbf{ + }}{{\mathbf{z}}_{{\mathbf{\alpha - \gamma }}}}{\mathbf{)/}}\sqrt {\mathbf{n}} \) .Show that w is minimized for the choice 蠏 =伪/2, so that the symmetric interval is the shortest. (Hints: (a) By definition of za, 蠒 (za) = 1 - 伪, so that za =蠒-1 (1 -伪 );

(b) the relationship between the derivative of a function y= f(x) and the inverse function x= f-1(y) is (dydy) f-1(y) = 1/f鈥�(x).)

For those of you who don鈥檛 already know, dragon boat racing is a competitive water sport that involves 20 paddlers propelling a boat across various race distances. It has become increasingly popular over the last few years. The article 鈥淧hysiological and Physical Characteristics of Elite Dragon Boat Paddlers鈥� (J. of Strength and Conditioning, 2013: 137鈥�145) summarized an extensive statistical analysis of data obtained from a sample of 11 paddlers. It reported that a 95% confidence interval for true average force (N) during a simulated 200-m race was (60.2, 70.6). Obtain a 95% prediction interval for the force of a single randomly selected dragon boat paddler undergoing the simulated race.

A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (229.764, 233.504). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval? (Hint: Use the center of the interval and its width to determine \(\overline x \) and s.)

When the population distribution is normal, the statistic median \(\left\{ {\left| {{{\rm{X}}_{\rm{1}}}{\rm{ - }}\widetilde {\rm{X}}} \right|{\rm{, \ldots ,}}\left| {{{\rm{X}}_{\rm{n}}}{\rm{ - }}\widetilde {\rm{X}}} \right|} \right\}{\rm{/}}{\rm{.6745}}\) can be used to estimate \({\rm{\sigma }}\). This estimator is more resistant to the effects of outliers (observations far from the bulk of the data) than is the sample standard deviation. Compute both the corresponding point estimate and s for the data of Example \({\rm{6}}{\rm{.2}}\).
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