Question

1.  Recall that the numbers ±1.96 cut off the center 95% of N(0, 1)

1. Use Minitab to create a graph showing the 90% confidence multiplier for the t with 63 degrees of freedom. Briefly explain, using what we have discussed about the t distributions, why this number is larger than 1.645.
2. Will the 95% confidence multiplier for the t with 63 degrees of freedom be larger or smaller than 1.96? (Hint: You may check this with Minitab.) Explain your answer using what we have discussed about the t distributions.

Answer 1

To create a 90% confidence interval for t with 63 degrees of freedom,

using minitab, we get critical value $t_{critical} = t_{0.05,63}=1.6694$

As degrees of freedom for t distribution increases, t distribution tends to normal distribution.

Normal distribution has higher frequency around the mean than t distribution, for t distribution, there is quite higher frequencies available around the tail, therefore

$P(t_{63}>t_0) = P(Z>1.645) = 0.05$ but t₀ will have higher value.

as $P(t_{63}>1.645) = 0.052475 > 0.05$

b)

To create a 95% confidence interval for t with 63 degrees of freedom,

using minitab, we get critical value $t_{critical} = t_{0.025,63}=1.99834$

So, it is larger than 1.96 for the same reason as mentioned in part a)