Suppose that the exam scores for students in a large university course are normally distributed with an unknown mean and standard deviation. The scores of 42 randomly sampled students in the course are used to estimate the mean of the population. What t-score should be used to find the 99% confidence interval for the population mean? df...4041424344t0.10…1.3031.3031.3021.3021.301t0.05…1.6841.6831.6821.6811.680t0.025…2.0212.0202.0182.0172.015t0.01…2.4232.4212.4182.4162.414t0.005…2.7042.7012.6982.6952.692 Use the portion of the table above or a calculator. If you use a calculator, round your answer to three decimal places.

Critical value t-score=2.701.

Step-by-step explanation:

When we calculate a confidence interval with an unknown population standard deviation, we estimate it from the sample standard deviation and use the t-score instead of the z-score.

The critical value for t depends on the level of confidence and the degrees of freedom.

If the sample size is 42, the degrees of freedom are:

For a confidence level of 99% and 41 degrees of freedom, the critical value of t is t=2.701.

Step-by-step explanation:

5x + 4y = 8   ⇒   4y = 8 - 5x  ⇒  y = (8 - 5x)÷4

2x − 3y = 17

2x - 3(8 - 5x)÷4 = 17    

          ×4            ×4    

8x - 3(8 - 5x) = 68

8x - 24 + 15x = 68

          +24       +24

           23x  =  92  

           ÷23      ÷23

               x = 4