Question 1039784
First use a calculator to find the t critical value.


Use the invT function on your TI83/TI84/etc calculator to get the result of -2.896 approximately (see screenshot below)


The 98% confidence interval will have 100-98 = 2% of the area in the tails. So 2/2 = 1% of the area is in the left tail. The goal is to find the value of k such that P(T < k) = 0.01. So that's why 0.01 is typed in as part of the calculator input (see screenshot below)


The degrees of freedom (df) is df = n-1 = 9-1 = 8. So that explains the 8 as the second input (see screenshot below)




<img src = "http://image.prntscr.com/image/68cbcadd9512496e9e7e3e39fb045525.png">


Note: to get the invT function, you hit the blue 2ND key. Then you hit the VARS key. The invT function is at #4.


The positive version of this number is the value we want. So the critical t value is approximately t = 2.896


If you do not have a TI calculator, such as a TI83 or TI84, then you can use a table like <a href="http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf">this one</a> (often found in the back of your textbook) to find the critical value. Look at the row df = 8. Then find the column that corresponds to the confidence level of 98%. The value 2.896 will be found at this row and column combo.


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Let's compute the lower bound and upper bound. We'll make L be the lower bound and U the upper bound.


We will use these formulas
*[Tex \LARGE L = \bar{x} - t*\frac{s}{\sqrt{n}}]


*[Tex \LARGE U = \bar{x} + t*\frac{s}{\sqrt{n}}]


In this case,
*[Tex \LARGE \bar{x} = 9.8] is the sample mean
s = 0.30 is the standard deviation
n = 9 is the sample size
t = 2.896 is the critical value we just found with the calculator

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Let's plug these values in and compute


First let's find the lower bound L
*[Tex \LARGE L = \bar{x} - t*\frac{s}{\sqrt{n}}]


*[Tex \LARGE L = 9.8 - 2.896*\frac{0.30}{\sqrt{9}}]


*[Tex \LARGE L = 9.5104]


*[Tex \LARGE L = 9.51]
So the lower bound is approximately 9.51


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Now let's find the upper bound U
*[Tex \LARGE U =  \bar{x} + t*\frac{s}{\sqrt{n}}]


*[Tex \LARGE U =  9.8 + 2.896*\frac{0.30}{\sqrt{9}}]


*[Tex \LARGE U =  10.0896]


*[Tex \LARGE U =  10.09]
So the upper bound is approximately 10.09


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To summarize, we found that
L = 9.51
U = 10.09


The 98% confidence interval is therefore (L,U) = <font color=red>(9.51, 10.09)</font>


So the answer is <font color=red>Choice B</font>