Question

3. It is desired to estimate the mean number of times a college student eats pizza in a week. A SRS of 16 college students was taken resulting in a sample mean of 2.54 and a sample standard deviation of 0.61. Assume this data is normally distributed. a. Give a 98% confidence interval for the true mean number of times a college student eats pizza in a week. b. Give a 98% confidence interval for the variance of the number of times a college student eats pizza in a week.

Answer 1

solution Given n=16 Ź S=0.61 X = 2.54 d 98% confidence interval for the true Construct mean. c.5= (xt texn-1) s =(2:54 + t(0.0,5) 062 let tla, nu)= tool,15) 2.602 a from table =(2:54† 2.602 (0:62 ) ) 2.602 =(2.54 – 0.3968 )) ( D = =( 2.1432 , 2.9368) Confidence interval is the (2.14, 2.94 )

construct 98% confidence interval for the variaora of the number of times a college students eats pizza in a week. (ni)s² 6.5 = 2 (ny) 5² c62c ne 6 s hy not? +le کار امسالاد(دمپ = 30.578 - from table X ca n-1) = flagg, is) = 5.229 from table = (164) (0:61) 2 (16-1) (0.61322 I 30.578 = 6 9.229 = 3 0.1825 582 € 1006743 = {0.18 € 6²5 1.07} Confidence interal is the