Question

The data below refers to the time in hours spent on mobile internet by sample of 10 students in a class. 39 42 47 45 32 45 37 34 33 29 Assume that the population data follows a normal distribution with unknown mean and unknown standard deviation. Find a 95% confidence interval estimate of μ . [33.813, 42.787] hours [35.557, 41.936] hours [34.664, 41.936] hours

Answer 1

SOLUTION:

From given data,

The data below refers to the time in hours spent on mobile internet by sample of 10 students in a class. 39 42 47 45 32 45 37 34 33 29 Assume that the population data follows a normal distribution with unknown mean and unknown standard deviation.

Sample size = n = 10

X	x-	( x- ) 2
39	39-38.3 = 0.7	(0.7)2 = 0.49
42	42-38.3 = 3.7	(3.7)2 = 13.69
47	47-38.3 = 8.7	(8.7)2 = 75.69
45	45-38.3 = 6.7	(6.7)2 = 44.89
32	32-38.3 = -6.3	(-6.3)2 = 39.69
45	45-38.3 = 6.7	(6.7)2 = 44.89
37	37-38.3 = -1.3	(-1.3)2 = 1.69
34	34-38.3 = -4.3	(-4.3)2 = 18.49
33	33-38.3 = -5.3	(-5.3)2 = 28.09
29	29-38.3 = -9.3	(-9.3)2 = 86.49
x = 383		( x- ) 2 = 354.1

x = 39 +42+ 47+ 45+ 32+ 45+ 37+ 34+ 33+ 29 =383

Mean = = x / n = 383 / 10 = 38.3

( x- ) ² =  0.49+13.69+75.69+44.89+39.69+44.89+1.69+18.49+28.09+86.49 = 354.1

Standard deviation = s = sqrt ( ( x- ) ² / (n-1)) = sqrt (354.1 / (10-1)) = sqrt (354.1 / 9) = 6.272

Degree of freedom (df) = n-1 = 10-1 = 9

Find a 95% confidence interval estimate of μ .

95% confidence interval

95% = 95/100 = 0.95

= 1-0.95 = 0.05

/2 = 0.05/2 = 0.025

Critical value:

t_/2,df = t_0.025,9 = 2.262157

The confidence interval :

- * ( / ) <   <   + * ( / )

38.3 - 2.262157 * (6.272 / ) <   <  38.3 + 2.262157 * (6.272 / )

38.3 - 4.48671819 <   <  38.3 + 4.48671819

33.813 <   <  42.787

95% confidence interval estimate of μ is [33.813, 42.787] hours.