Chapter 23, Problem 2E

(a)

To determine

To estimate the critical value of t for a $95\%$ confidence interval with $\text{df}=7$ , using the t table, software or a calculator.

Answer to Problem 2E

The t value for a $95\%$ confidence interval with $\text{df}=7$ is $2.365$ .

Explanation of Solution

Now, we have to calculate the t value for a $95\%$ confidence interval with $\text{df}=7$ so, we will use excel to calculate the value. Thus,

Formula used:

TINV function returns the two-tailed inverse of the Student's t -distribution. Thus, the syntax for the same is as:

TINV ( $2\times$ probability, deg_freedom)

Calculation:

Now, it is given that:

It is a one tail test,

  $\begin{array}{l}\text{df}=7\\ c=0.95\\ \frac{\alpha }{2}=\frac{1-0.95}{2}\\ =\frac{0.05}{2}\\ =0.025\end{array}$

Now, by putting above information in the formula we have,

t-value

=TINV(2*0.025,7)

Now, after calculating it we have,

t-value

2.365

Thus, we have that the t value for a $95\%$ confidence interval with $\text{df}=7$ is $2.365$ .

(b)

To determine

To estimate the critical value of t for a $99\%$ confidence interval with $\text{df}=102$ , using the t table, software or a calculator.

Answer to Problem 2E

The t value for a $99\%$ confidence interval with $\text{df}=102$ is $2.625$ .

Explanation of Solution

Now, we have to calculate the t value for a $99\%$ confidence interval with $\text{df}=102$ so, we will use excel to calculate the value. Thus,

Formula used:

TINV function returns the two-tailed inverse of the Student's t -distribution. Thus, the syntax for the same is as:

TINV ( $2\times$ probability ,deg_freedom)

Calculation:

Now, it is given that:

It is a one tail test,

  $\begin{array}{l}\text{df}=102\\ c=0.99\\ \frac{\alpha }{2}=\frac{1-0.99}{2}\\ =\frac{0.01}{2}\\ =0.005\end{array}$

Now, by putting above information in the formula we have,

t-value

=TINV(2*0.005,102)

Now, after calculating it we have,

t-value

2.625

Thus, we have that the t value for a $99\%$ confidence interval with $\text{df}=102$ is $2.625$ .

(c)

To determine

To estimate the P-value for $t\le 2.19$ with $41$ degrees of freedom, using the t table, software or a calculator.

Answer to Problem 2E

The P-value for $t\le 2.19$ with $41$ degrees of freedom be $0.98287$ .

Explanation of Solution

Now, we have to calculate the P-value for $t\le 2.19$ with $41$ degrees of freedom so, we will use excel to calculate the value. Thus,

Formula used:

TDIST returns the Percentage Points (probability) for the Student t -distribution where a numeric value ( x ) is a calculated value of t for which the Percentage Points are to be computed. Thus, the syntax for the same is as:

TDIST ( x , deg_freedom, tails)

Calculation:

Now, it is given that:

It is a one tail test,

  $\begin{array}{l}\text{df}=41\\ x=2.19\end{array}$

Now, by putting above information in the formula we have,

P-value

=1-TDIST(2.19,41,1)

Now, after calculating it we have,

P-value

0.98287

Thus, we have that the P-value ffor $t\le 2.19$ with $41$ degrees of freedom be $0.98287$ .

(d)

To determine

To estimate the P-value for $\left|t\right|>2.33$ with $12$ degrees of freedom, using the t table, software or a calculator.

Answer to Problem 2E

The P-value for $\left|t\right|>2.33$ with $12$ degrees of freedom be $0.03807$ .

Explanation of Solution

Now, we have to calculate the P-value for $\left|t\right|>2.33$ with $12$ degrees of freedom so, we will use excel to calculate the value. Thus,

Formula used:

TDIST returns the Percentage Points (probability) for the Student t -distribution where a numeric value ( x ) is a calculated value of t for which the Percentage Points are to be computed. Thus, the syntax for the same is as:

TDIST ( x , deg_freedom, tails)

Calculation:

Now, it is given that:

It is a two tail test,

  $\begin{array}{l}\text{df}=12\\ \left|t\right|>2.33\end{array}$

Now, by putting above information in the formula we have,

P-value

=TDIST(2.33,12,2)

Now, after calculating it we have,

P-value

0.03807

Thus, we have that the P-value for $\left|t\right|>2.33$ with $12$ degrees of freedom be $0.03807$ .