Question

What critical value would you use for a $95\%$ confidence interval based on $t\left( 21 \right)$ distribution?

Answer 1

Hint: This type of question needs the knowledge of test distribution. The t-value measures the size of the difference relative to the variation in your sample data. To solve the problem we will first find the significance level with the help of confidence percentage.

Complete step-by-step answer:
The question asks us to find the critical value that needs to be used for $95\%$ confidence interval which is based on the $t\left( 21 \right)$ distribution. Starting with the explanation of t-distribution. T-tests are statistical hypothesis test that you use to analyze one or two sample means. The T distribution is a continuous probability distribution of the z-score when the estimated standard deviation is used in the denominator rather than the true standard deviationThe first step is to find the significance level which is the difference between $1$ and confidence. So mathematically it will be represented as:
Significance level = $\alpha $ = $\text{1- confidence}$
The value of the confidence given in the question is $95\%$ . We need to convert the percentage into decimal form which could be done by dividing the number by $100$.
$\Rightarrow \dfrac{95}{100}=0.95$
On putting the decimal value we get:
$\Rightarrow \text{1- 0}\text{.95}$
$\Rightarrow 0.05$
The second step is to find the critical t- value = ${{t}_{\dfrac{\alpha }{2}}}$
$\Rightarrow {{\text{t}}_{\dfrac{0.05}{2}}}$
\[\Rightarrow {{\text{t}}_{0.025}}\]
The below table entries represent the area under the standard normal curve.

dt/p	0.10	0.05	0.025	0.01	0.0005
19	1.327728	1.729133	2.09302	2.53948	2.86093
20	1.325431	1.724718	2.08596	2.52798	2.84534
21	1.323188	1.720743	2.07961	2.51765	2.83136
22	1.321237	1.717144	2.07387	2.50832	2.81876
23	1.319460	1.713872	2.06866	2.49987	2.80734

As in the question, df is given as $21$ and the confidence critical value is $0.025$, so on taking the intersection of the two we get $2.07961$ .
$\therefore $ The critical value that would be used for a $95\%$ confidence interval based on $t\left( 21 \right)$ distribution is $2.07961$.

Note: To convert the percentage into the fraction we actually divide the number in fraction by hundred removing the percentage sign. For example $32\%$ can be converted into fraction by $\dfrac{32}{100}$. The T-distribution, like the normal distribution, is bell-shaped and symmetric, but it has heavier tails, which means it tends to produce values that fall far from its mean. T-tests are used in statistics to estimate significance.