Chapter 23, Problem 1E

(a)

To determine

To estimate the critical value of t for a $90\%$ confidence interval with $\text{df}=17$ , using the t table, software or a calculator.

Answer to Problem 1E

The t value for a $90\%$ confidence interval with $\text{df}=17$ is $1.74$ .

Explanation of Solution

Now, we have to calculate the t value for a $90\%$ confidence interval with $\text{df}=17$ 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,

  $\begin{array}{l}\text{df}=17\\ c=0.90\\ \frac{\alpha }{2}=\frac{1-0.90}{2}\\ =\frac{0.10}{2}\\ =0.05\end{array}$

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

t-value

=TINV(2*0.05,17)

Now, after calculating it we have,

t-value

1.740

Thus, we have that the t value for a $90\%$ confidence interval with $\text{df}=17$ is $1.74$ .

(b)

To determine

To estimate the critical value of t for a $98\%$ confidence interval with $\text{df}=88$ , using the t table, software or a calculator.

Answer to Problem 1E

The t value for a $98\%$ confidence interval with $\text{df}=88$ is $2.37$ .

Explanation of Solution

Now, we have to calculate the t value for a $98\%$ confidence interval with $\text{df}=88$ 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,

  $\begin{array}{l}\text{df}=88\\ c=0.98\\ \frac{\alpha }{2}=\frac{1-0.98}{2}\\ =\frac{0.02}{2}\\ =0.01\end{array}$

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

t-value

=TINV(2*0.01,88)

Now, after calculating it we have,

t-value

2.37

Thus, we have that the t value for a $98\%$ confidence interval with $\text{df}=88$ is $2.37$ .

(c)

To determine

To estimate the P-value for $t\ge 2.09$ with $4$ degrees of freedom, using the t table, software or a calculator.

Answer to Problem 1E

The P-value for $t\ge 2.09$ with $4$ degrees of freedom be $0.052415$ .

Explanation of Solution

Now, we have to calculate the P-value for $t\ge 2.09$ with $4$ degrees of freedomso, 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,

  $\begin{array}{l}\text{df}=4\\ x=2.09\end{array}$

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

P-value

=TDIST(2.09,4,1)

Now, after calculating it we have,

P-value

0.052415

Thus, we have that the P-value for $t\ge 2.09$ with $4$ degrees of freedom be $0.052415$ .

(d)

To determine

To estimate the P-value for $\left|t\right|>1.78$ with $22$ degrees of freedom, using the t table, software or a calculator.

Answer to Problem 1E

The P-value for $\left|t\right|>1.78$ with $22$ degrees of freedom be $0.088894$ .

Explanation of Solution

Now, we have to calculate the P-value for $\left|t\right|>1.78$ with $22$ 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,

  $\begin{array}{l}\text{df}=22\\ \left|t\right|>1.78\end{array}$

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

P-value

=TDIST(1.78,22,2)

Now, after calculating it we have,

P-value

0.088894

Thus, we have that the P-value for $\left|t\right|>1.78$ with $22$ degrees of freedom be $0.088894$ .