Now suppose we have n = 25 and we know the heights of the men do follow the Normal Curve. The sample average = 70" and sample SD = 3.3". Let's compare a 90% CI using SE_avgand z^*to a 90% CI using SE⁺_avgand t^*.

Using Z:

1. SE_avg=0.66"
2. Computer's answer now shown above.You are correct.
3. Your receipt no. is155-1592Previous Tries
4. z^*for 90% CI =1.65
5. Computer's answer now shown above.You are correct.
6. Your receipt no. is155-6789Previous Tries
7. 90% CI = (68.91" to71.09")
8. Computer's answer now shown above.You are correct.
9. Your receipt no. is155-4687Previous Tries

Using t:

1. SE⁺_avg="
2. Incorrect.Tries 2/3Previous Tries
3. df =24
4. Computer's answer now shown above.You are correct.
5. Your receipt no. is155-7793Previous Tries
6. t^*for 90% CI = ____1.71
7. Hint:The t-table in your notes gives the critical values of t corresponding to the areas of right-hand tails for different degrees of freedom, but you want the critical values of t corresonding to middle areas (confidence intervals). For example, a right-hand tail of 2.5% would correspond to a CI interval of 95%.Computer's answer now shown above.You are correct.
8. Your receipt no. is155-2747Previous Tries
9. 90% CI = (" to")
10. Incorrect.Tries 3/5Previous Tries

g.Confidence Intervals using t will always be _______ than those using Z.

Correct:wider

Incorrectnarrower

Incorrectdepends on the degrees of freedom