Suppose that the annual interest rates on mortgages in the U.S. are normally distributed with an unknown mean and standard deviation. The rates of 22 randomly sampled mortgages (home loans) are used to estimate the mean of the population. What t-score should be used to find the 98% confidence interval for the population mean?

Respuesta :

Answer:

The value is [tex]t_{\frac{\alpha }{2} , 21 } =  2.518[/tex]

Step-by-step explanation:

From the question we are told that

   The  sample size is  n =  22

Generally the degree of freedom is mathematically represented as

       [tex]df = n- 1[/tex]

=>    [tex]df = 22- 1[/tex]

=>    [tex]df = 21[/tex]

From the question we are told the confidence level is  98% , hence the level of significance is    

      [tex]\alpha = (100 - 98 ) \%[/tex]

=>   [tex]\alpha = 0.02[/tex]

Generally from the t distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex]  at a degree of freedom of [tex]df = 21[/tex]  is  

   [tex]t_{\frac{\alpha }{2} , 21 } =  2.518[/tex]

This is right and I just need these points for Brainly