Can you explain to me how to find alpha of t with a confiden

18, A sample of size n = 44 has sample mean X = 56 9 and sample standard deviation s-91 a. Construct a 98% confidence interval for the population mean b. If the sample size were n = 30, would the confidence interval be narrower or wider? 19. A sample of size n 89 has sample mean 87.2 and sample standard deviation s 53 a. Construct a 95% confidence interval tor the population mean Answer b. If the confidence level were 99%, would the confidence interval be narrower or wider? Answer 20, A sample of size n = 35 has sample mean = 3485 and sample standard deviation s = 17.9 a. Construct a 98% confidence interval for the population mean b. If the confidence level were 95%, would the confidence interval be narrower or wider? Working with the Concepts 21. Online courses: A sample of 263 students who were taking online courses were asked to describe their overall impression of online learning on a scale of 1-7, with 7 representing the mos favorable impression. The average score was 5 53, and the standard deviation was 0.92 a. Construct a 95% confidence interval for the mean score. Answer b. Assume that the mean score for students taking traditional courses is 5 55 A college that offers online courses claims that the mean scores for online courses and traditional course

Solution

Confidence Interval = sample mean (+-) t _{( n-1,alpha/2 )} ( s / sqrt(n) )

18. (a) n =44 , sample mean = 56.9 , s =9.1 , alpha = 0.02 ,  t _{( n-1,alpha/2 )} = 2.423

the confidence interval is ( 53.576 , 60.224 ) [ The difference 60.224 - 53.576 = 6.648 ]

( b ) for n= 30 , sample mean = 56.9 , s =9.1 ,alpha = 0.02 , t _{( n-1,alpha/2 )} = 2.462

the confidence interval is ( 52.8096 , 60.9904 ) [ The difference 60.9904 - 52.8096 = 8.1808 ]

The width of the interval of (b) is bigger than (a) , so we can say that with the decrease in the value of n the interval becomes wider.

19. (a) n=89, sample mean=87.2, s = 5.3 , alpha = 0.05 , t _{( n-1,alpha/2 )} = 1.990

the confidence interval is ( 86.0821 , 88.3179 ) [ The difference 88.3179 - 86.0821 = 2.2358 ]

(b) n=89, sample mean=87.2, s = 5.3 , alpha =0.01 , t _{( n-1,alpha/2 )} = 2.639

the confidence interval is ( 85.7175 , 88.6825 ) [ The difference 88.6825 - 85.7175 = 2.965 ]

The interval becomes wider.

20. (a) n=35 , sample mean =34.85 , s=17.9 , alpha = 0.02 , t _{( n-1,alpha/2 )} = 2.457

the confidence interval is ( 27.4159 , 42.2841 ) [ The difference 42.2841- 27.4159 = 14.8682 ]

(b) n=35 , sample mean =34.85 , s=17.9 , alpha = 0.05 , t _{( n-1,alpha/2 )} = 2.042

the confidence interval is ( 28.6716 , 41.0284 ) [ The difference 41.0284 - 28.6716 = 12.3568 ]

the interval becomes narrower.