Answer: 1.745884
Step-by-step explanation:
Given : Level of significance : [tex]1-\alpha:0.90[/tex]
Then , significance level : [tex]\alpha: 1-0.90=0.10[/tex]
Sample size : [tex]n=17[/tex]
Then , the degree of freedom for t-distribution: [tex]df=n-1=16[/tex]
The t-score for a level of confidence [tex](1-\alpha)[/tex] is given by the formula below :-
[tex]t_{(df,\alpha/2)}[/tex], where df is the degree of freedom and [tex]\alpha[/tex] is the significance level.
With the help of the normal t-distribution table, we have
[tex]\text{ t-score}=t_{(df,\alpha/2)}=t_{16,0.05}=1.745884[/tex]
Thus, the t-score should be used to find the 90% confidence interval for the population mean =1.745884
