95% confidence interval help

#1
How much money does the average professional football fan spend on food at a single football game? That question was posed to ten randomly selected football fans. The sampled result showed that the sample mean and standard deviation were $70.00 and $17.50 respectively. Use this information to create a 95 percent confidence interval for the population mean.


using the zchart I came up with this so far

70+- 1.960(17.50/v-10) (square root of 10)

somehow this is not right, 1.960 should actually be 2.262 and I'm not sure how, any help would be appreciated.
 

JohnM

TS Contributor
#2
Because you need to use the t distribution, not the normal distribution. The t statistic for 95% confidence with 9 d.o.f. (10 - 1) is 2.262
 

JohnM

TS Contributor
#4
As a rule of thumb, when the sample size is less than 30 (i.e., "small"), the sampling distribution of the mean more closely follows a t distribution than a normal distribution.
 
#5
ahhh gotcha, I'm looking the t distribution now and see the 2.262. So the sample size of 10 is what determined to use t distribution? so if the sample size was > than 30 I would have gotten my original answer of 1.960?
 
#7
also, if my sample size was say 20 which would be (dof 19) it would have been 2.093?

One other question, why would the chart go above 30 if you would use the z if the sample size was greater than 30? Thank you so much for your help btw, I've been trying to figure htis out the past couple hours
 

JohnM

TS Contributor
#8
sinful said:
also, if my sample size was say 20 which would be (dof 19) it would have been 2.093?
yes

sinful said:
One other question, why would the chart go above 30 if you would use the z if the sample size was greater than 30? Thank you so much for your help btw, I've been trying to figure htis out the past couple hours
Theoretically, as n increases, the t distribution "approaches" a normal distribution but never really "gets there" exactly. When n>30, we say that it's "close enough."