Answer:
Could you give the question in details please?
Answer:
a. t*
b. (22.87, 24.53)
c. yes
Step-by-step explanation:
We have the following data:
mean = m = 23.7
standard deviation = sd = 2.15
x = 20
Thus:
a. The critical value would be t * since the value is m <30
b. whenever the mean is less than 30, the t distribution is used, for 95% confidence interval with degree of freedom = (x - 1) df
= 20 - 1
= 19
now, for this value we look in the table of t, we find that t * = 1.729
m + - t * (sd / x ^ 81/2))
replacing
23.7 + - 1.729 * (2.15 / 20 ^ (1/2))
23.7 + - 0.8312
then the inverval would be:
23.7 + 0.8312 = 24.5312
23.7 - 0.8312 = 22.8688
(22.87, 24.53)
c. yes, the interval is accurate since the mean of 23.7 is within this value
5920 that is the result of 5.92 by 10 to the third power