Question

For parts a and bâ, use technology to estimate the following. âa) The critical value of t for a 90â% confidence interval with dfequals3. âb) The critical value of t for a 99â% confidence interval with dfequals106.

Answer 1

The critical value of t for a 90% confidence interval with df=3 is approximately 3.182. The critical value of t for a 99% confidence interval with df=106 is approximately 2.626.

Step-by-step explanation:

a) To find the critical value of t for a 90% confidence interval with df=3, you can use a t-distribution table or a calculator. Using a table or calculator, the critical value can be found as approximately 3.182.
b) For a 99% confidence interval with df=106, the critical value of t can be found using a t-distribution table or a calculator. The value is approximately 2.626.

Answer 2

a) For the 90% confidence interval the value of
`\alpha=1-0.9=0.1` and
`\alpha/2=0.05`, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

`t_(\alpha/2) =\pm 2.35`

b) For the 99% confidence interval the value of
`\alpha=1-0.99=0.01` and
`\alpha/2=0.005`, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

`t_(\alpha/2) =\pm 2.62`

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

Part a

For the 90% confidence interval the value of
`\alpha=1-0.9=0.1` and
`\alpha/2=0.05`, with that value we can find the quantile required for the interval in the t distribution with df =3. And we can use the folloiwng excel code: "=T.INV(0.05,3)" and we got:

`t_(\alpha/2) =\pm 2.35`

Part b

For the 99% confidence interval the value of
`\alpha=1-0.99=0.01` and
`\alpha/2=0.005`, with that value we can find the quantile required for the interval in the t distribution with df =106. And we can use the folloiwng excel code: "=T.INV(0.005,106)" and we got:

`t_(\alpha/2) =\pm 2.62`