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Chapter 7: Q31E (page 302)

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 CI is desired for the true average stray-load loss \({\rm{\mu }}\) (watts) for a certain type of induction motor when the line current is held at \({\rm{10 amps}}\) for a speed of \({\rm{1500 rpm}}\). Assume that stray-load loss is normally distributed with \({\rm{\sigma = 3}}{\rm{.0}}\). a. Compute a \({\rm{95\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 25}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). b. Compute a \({\rm{95\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). c. Compute a \({\rm{99\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). d. Compute an \({\rm{82\% }}\) CI for \({\rm{\mu }}\) when \({\rm{n = 100}}\) and \({\rm{\bar x = 58}}{\rm{.3}}\). e. How large must n be if the width of the \({\rm{99\% }}\) interval for \({\rm{\mu }}\) is to be \({\rm{1}}{\rm{.0}}\)?

presented a sample of \({\rm{n = 153}}\)observations on ultimate tensile strength, the previous section gave summary quantities and requested a large-sample confidence interval. Because the sample size is large, no assumptions about the population distribution are required for the validity of the CI.

a. Is any assumption about the tensile-strength distribution required prior to calculating a lower prediction bound for the tensile strength of the next specimen selected using the method described in this section? Explain.

b. Use a statistical software package to investigate the plausibility of a normal population distribution.

c. Calculate a lower prediction bound with a prediction level of \({\rm{95\% }}\)for the ultimate tensile strength of the next specimen selected.

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.)

It is important that face masks used by fire fighters be able to withstand high temperatures because fire fighters commonly work in temperatures of 200–500°F. In a test of one type of mask, 11 of 55 masks had lenses pop out at 250°. Construct a 90% upper confidence bound for the true proportion of masks of this type whose lenses would pop out at 250°.

The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. The article “Testing Practices for the AC Breakdown Voltage Testing of Insulation Liquids'' (IEEE)gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions.

\(\begin{array}{l}{\rm{62\;50\;53\;57\;41\;53\;55\;61\;59\;64\;50\;53\;64\;62}}\\{\rm{\;50\;68 54\;55\;57\;50\;55\;50\;56\;55\;46\;55\;53\;54\;}}\\{\rm{52\;47\;47\;55 57\;48\;63\;57\;57\;55\;53\;59\;53\;52\;}}\\{\rm{50\;55\;60\;50\;56\;58}}\end{array}\)

a. Construct a boxplot of the data and comment on interesting features.

b. Calculate and interpret a \({\rm{95\% }}\)CI for true average breakdown voltage m. Does it appear that m has been precisely estimated? Explain.

c. Suppose the investigator believes that virtually all values of breakdown voltage are between \({\rm{40 and 70}}\). What sample size would be appropriate for the \({\rm{95\% }}\)CI to have a width of 2 kV (so that m is estimated to within 1 kV with \({\rm{95\% }}\)confidence)
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