Question

"Suppose that you are a city planner who obtains a sample of 20 randomly selected members of a mid-sized town in order to determine the average amount of money that residents spend on transportation each month (such as fuel, vehicle repairs, and public transit). To 3 decimal places, what is the critical value for the 95% confidence interval

Answer 1

Answer: 2.093

Step-by-step explanation:

As per give , we have

Sample size : n= 20

Degree of freedom : df= n-1=19

Significance level : $\alpha: 1-0.95=0.05$

Since , the sample size is small (n<30) so we use t-test.

For confidence interval , we find two-tailed test value.

Using students's t-critical value table,

Critical t-value : $t_{\alpha/2, df}=t_{0.025,19}=2.093$

Thus, the critical value for the 95% confidence interval = 2.093