Suppose that the hourly wages of fast food workers are normally distributed with an unknown mean and standard deviation. the wages of 40 randomly sampled fast food workers are used to estimate the mean wage of the population. what t-score should be used to find the 80% confidence interval for the population mean?

1.303639

Step-by-step explanation:

The t-score for a level of confidence is given by :_

, where df is the degree of freedom and is the significance level.

Given : Level of significance :

Then , significance level :

Sample size :

Then , the degree of freedom for t-distribution:

Using the normal t-distribution table, we have

Thus, the t-score should be used to find the 80% confidence interval for the population mean =1.303639

use this formula:

p(a\b) = p(a and b)/p(b).

take it from here.

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)