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

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

Short Answer

Expert verified

The t critical value for a lower or an upper confidence bound

\(\begin{array}{l}{\rm{a}}{\rm{. }}{{\rm{t}}_{{\rm{0}}{\rm{.05,10}}}}{\rm{ - 1}}{\rm{.812;}}\\{\rm{b}}{\rm{. }}{{\rm{t}}_{{\rm{0}}{\rm{.05,15}}}}{\rm{ - 1}}{\rm{.753; }}\\{\rm{c}}{\rm{. }}{{\rm{t}}_{{\rm{0}}{\rm{.01,15}}}}{\rm{ - 2}}{\rm{.602; }}\\{\rm{d}}{\rm{. }}{{\rm{t}}_{{\rm{0}}{\rm{.01,4}}}}{\rm{ - 3}}{\rm{.747;}}\\{\rm{e}}{\rm{. }}{{\rm{t}}_{{\rm{0}}{\rm{.02,24}}}}{\rm{ - 2}}{\rm{.172; }}\\{\rm{f}}{\rm{. }}{{\rm{t}}_{{\rm{0}}{\rm{.01,37}}}}{\rm{ - 2}}{\rm{.429}}\end{array}\)

Step by step solution

01

Step 1: \({\rm{Confidence level  = 95\% , df  = 10}}\)

Upper confidence bound for \({\rm{\mu }}\)and lower confidence bound for \({\rm{\mu }}\)are given by

\(\begin{array}{l}{\rm{\bar x + }}{{\rm{t}}_{{\rm{\alpha ,n - 1}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ upper bound, }}\\{\rm{\bar x - }}{{\rm{t}}_{{\rm{\alpha ,n - 1}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ lower bound, }}\end{array}\)

with confidence level of \({\rm{100(1 - \alpha )\% }}\)The distribution the random sample is taken from is normal.

(a):

The critical value \({{\rm{t}}_{{\rm{a,n - 1}}}}\)for an upper/lower confidence bound with confidence level\(95\% \), can be obtained as

\(\begin{array}{*{20}{r}}{{\rm{100(1 - \alpha ) - 95}}}\\{{\rm{\alpha - 0}}{\rm{.05}}}\end{array}\)

and, from the table A. 5 in the appendix, for \(10\)degrees of freedom, and \({\rm{\alpha - 0}}{\rm{.05}}\)the t critical value is

\({{\rm{t}}_{{\rm{0}}{\rm{.05,10}}}}{\rm{ = 1}}{\rm{.812}}{\rm{.}}\)The area to the left of the critical value is shaded blue on the picture and has surface area of \({\rm{1 - \alpha - 0}}{\rm{.95}}\).

02

Step 2: \({\rm{ Confidence level  =  95\% , df  = 15}}\)

(b):

The critical value \({{\rm{t}}_{{\rm{\alpha ,n - 1}}}}\)for an upper/lower confidence bound, with confidence level \(95\% \), can be obtained as

\(\begin{array}{*{20}{r}}{{\rm{100(1 - \alpha ) - 95}}}\\{{\rm{\alpha - 0}}{\rm{.05}}}\end{array}\)

and, from the table A.5 in the appendix, for 15 degrees of freedom, and \({\rm{\alpha - 0}}{\rm{.05}}\)the t critical value is

\({{\rm{t}}_{{\rm{0}}{\rm{.05,15}}}}{\rm{ = 1}}{\rm{.753}}\)

The area to the left of the critical value is shaded blue on the picture and has surface area of \({\rm{1 - \alpha - 0}}{\rm{.95}}\).

03

Step 3: \({\rm{ Confidence level  =   99\% , df = 15}}\)

(c):

The critical value \({{\rm{t}}_{{\rm{\alpha ,n - 1}}}}\) for an upper/lower confidence bound, with confidence level \(99\% \), can be obtained as

\(\begin{array}{*{20}{r}}{{\rm{100(1 - \alpha ) - 99}}}\\{{\rm{\alpha - 0}}{\rm{.01}}}\end{array}\)

and, from the table A.5 in the appendix, for \(15\) degrees of freedom, and \({\rm{\alpha - 0}}{\rm{.01}}\) the t critical value is

\({{\rm{t}}_{{\rm{0}}{\rm{.01,15}}}}{\rm{ = 2}}{\rm{.602}}\)

The area to the left of the critical value is shaded blue on the picture and has surface area of \({\rm{1 - \alpha - 0}}{\rm{.99}}\).

04

Step 4: \({\rm{Confidence level  =   99\% , n = 5}}\)

(d):

The critical value \({{\rm{t}}_{{\rm{a,n - 1}}}}\)for an upper/lower confidence bound, with confidence level \(99\% \), can be obtained as

\(\begin{array}{*{20}{r}}{{\rm{100(1 - \alpha ) - 99}}}\\{{\rm{\alpha - 0}}{\rm{.01}}}\end{array}\)

and, from the table A. 5 in the appendix, for \({\rm{n = 1 - 4}}\)degrees of freedom, and \({\rm{\alpha - 0}}{\rm{.01}}\) the t critical value is

\({{\rm{t}}_{{\rm{0}}{\rm{.01,4}}}}{\rm{ = 3}}{\rm{.747}}\)

The area to the leftof the critical value is shaded blue on the picture and has surface area of \({\rm{1 - \alpha - 0}}{\rm{.99}}\)

05

Step 5: \({\rm{\; Confidence level  = 98\% , df =   24}}\)

(e):

The critical value \({{\rm{t}}_{{\rm{a,n - 1}}}}\) for an upper/lower confidence bound, with confidence level \(98\% \), can be obtained as

\(\begin{array}{*{20}{r}}{{\rm{100(1 - \alpha ) - 98}}}\\{{\rm{\alpha - 0}}{\rm{.02}}}\end{array}\)

and, from the table A.5 in the appendix, for \(24\) degrees of freedom, and \({\rm{\alpha - 0}}{\rm{.02}}\)$ the t critical value is

\({{\rm{t}}_{{\rm{0}}{\rm{.172,24}}}}{\rm{ = 2}}{\rm{.172}}\)

The area to the left of the critical value is shaded blue on the picture and has surface area of \({\rm{1 - \alpha - 0}}{\rm{.98}}\).

06

Step 6: \({\rm{ Confidence level  =   99\% , n = 38}}\)

(f)

The critical value \({{\rm{t}}_{{\rm{a,n - 1}}}}\) for an upper/lower confidence bound, with confidence level \(99\% \)can be obtained as

\(\begin{array}{*{20}{r}}{{\rm{100(1 - \alpha ) - 99}}}\\{{\rm{\alpha - 0}}{\rm{.01}}}\end{array}\)

and, from the table A.5 in the appendix, for \({\rm{38 - 1 = 37}}\)degrees of freedom, and \({\rm{\alpha - 0}}{\rm{.01}}\)the t critical value is

\({{\rm{t}}_{{\rm{0}}{\rm{.01,37}}}}{\rm{ - 2}}{\rm{.429}}\)

The area to the left of the critical value is shaded blue on the picture and has surface area of \({\rm{1 - \alpha - 0}}{\rm{.99}}\).

Hence

\(\begin{array}{l}{\rm{a}}{\rm{. }}{{\rm{t}}_{{\rm{0}}{\rm{.05,10}}}}{\rm{ - 1}}{\rm{.812;}}\\{\rm{b}}{\rm{. }}{{\rm{t}}_{{\rm{0}}{\rm{.05,15}}}}{\rm{ - 1}}{\rm{.753; }}\\{\rm{c}}{\rm{. }}{{\rm{t}}_{{\rm{0}}{\rm{.01,15}}}}{\rm{ - 2}}{\rm{.602; }}\\{\rm{d}}{\rm{. }}{{\rm{t}}_{{\rm{0}}{\rm{.01,4}}}}{\rm{ - 3}}{\rm{.747;}}\\{\rm{e}}{\rm{. }}{{\rm{t}}_{{\rm{0}}{\rm{.02,24}}}}{\rm{ - 2}}{\rm{.172; }}\\{\rm{f}}{\rm{. }}{{\rm{t}}_{{\rm{0}}{\rm{.01,37}}}}{\rm{ - 2}}{\rm{.429}}\end{array}\)

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Most popular questions from this chapter

A more extensive tabulation of t critical values than what appears in this book shows that for the t distribution with

\({\rm{20}}\)df, the areas to the right of the values \({\rm{.687, }}{\rm{.860, and 1}}{\rm{.064 are }}{\rm{.25, }}{\rm{.20, and }}{\rm{.15,}}\)respectively. What is the confidence level for each of the following three confidence intervals for the mean m of a normal population distribution? Which of the three intervals would you recommend be used, and why?

\(\begin{array}{l}{\rm{a}}{\rm{.(\bar x - }}{\rm{.687s/}}\sqrt {{\rm{21}}} {\rm{,\bar x + 1}}{\rm{.725s/}}\sqrt {{\rm{21}}} {\rm{)}}\\{\rm{b}}{\rm{.(\bar x - }}{\rm{.860\;s/}}\sqrt {{\rm{21}}} {\rm{,\bar x + 1}}{\rm{.325s/}}\sqrt {{\rm{21}}} {\rm{)}}\\{\rm{c}}{\rm{.(\bar x - 1}}{\rm{.064s/}}\sqrt {{\rm{21}}} {\rm{,\bar x + 1}}{\rm{.064s/}}\sqrt {{\rm{21}}} {\rm{)}}\end{array}\)

Exercise 72 of Chapter 1 gave the following observations on a receptor binding measure (adjusted distribution volume) for a sample of 13 healthy individuals: 23, 39, 40, 41, 43, 47, 51, 58, 63, 66, 67, 69, 72.

a. Is it plausible that the population distribution from which this sample was selected is normal?

b. Calculate an interval for which you can be 95% confident that at least 95% of all healthy individuals in the population have adjusted distribution volumes lying between the limits of the interval.

c. Predict the adjusted distribution volume of a single healthy individual by calculating a 95% prediction interval. How does this interval’s width compare to the width of the interval calculated in part (b)?

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 \({\rm{6}}{\rm{.2}}\).

Consider a normal population distribution with the value of \({\rm{\sigma }}\) known. a. What is the confidence level for the interval \({\rm{\bar x \pm 2}}{\rm{.81\sigma /}}\sqrt {\rm{n}} \)? b. What is the confidence level for the interval \({\rm{\bar x \pm 1}}{\rm{.44\sigma /}}\sqrt {\rm{n}} \)? c. What value of \({{\rm{z}}_{{\rm{\alpha /2}}}}\) in the CI formula (\({\rm{7}}{\rm{.5}}\)) results in a confidence level of \({\rm{99}}{\rm{.7\% }}\)? d. Answer the question posed in part (c) for a confidence level of \({\rm{75\% }}\).

In Example 6.8, we introduced the concept of a censored experiment in which n components are put on test and the experiment terminates as soon as r of the components have failed. Suppose component lifetimes are independent, each having an exponential distribution with parameter λ. Let Y1 denote the time at which the first failure occurs, Y2 the time at which the second failure occurs, and so on, so that Tr= Y1 + … + Yr+ (n- r)Yr is the total accumulated lifetime at termination. Then it can be shown that 2λ°Õr has a chi-squared distribution with 2r df. Use this fact to develop a 100(1 - α)% CI formula for true average lifetime \(\frac{{\bf{1}}}{{\bf{\lambda }}}\) . Compute a 95% CI from the data in Example 6.8.

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