Answer:
Explanation:
The question is incomplete. The complete question is:
It is known that the number of hours a student sleeps per night has a normal distribution. The sleeping time in hours of a random sample of 8 students is given below. 7.4, 6.2, 8.5, 6.3, 5.4, 5.5, 6.3, 8.3 Compute a 98% confidence interval for the true mean time a student sleeps per night and fill in the blanks appropriately. We have 98% confidence that the true mean time a student sleeps per night is between _____ and ____ hours. (Keep 3 decimal places)
Solution:
Mean = (7.4 + 6.2 + 8.5 + 6.3 + 5.4 + 5.5 + 6.3 + 8.3)/8 = 6.7375
Standard deviation = √(summation(x - mean)²/n
Summation(x - mean)² = (7.4 - 6.7375)^2 + (6.2 - 6.7375)^2 + (8.5 - 6.7375)^2 + (6.3 - 6.7375)^2 + (5.4 - 6.7375)^2 + (5.5 - 6.7375)^2 + (6.3 - 6.7375)^2 + (8.3 - 6.7375)^2 = 9.97875
Standard deviation = √(9.97875/8
s = 1.12
Confidence interval is written in the form,
(Sample mean - margin of error, sample mean + margin of error)
The sample mean, x is the point estimate for the population mean.
Margin of error = z × s/√n
Where
sample standard deviation
number of samples
From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score
In order to use the t distribution, we would determine the degree of freedom, df for the sample.
df = n - 1 = 8 - 1 = 7
Since confidence level = 98% = 0.98, α = 1 - CL = 1 - 0.98 = 0.02
α/2 = 0.02/2 = 0.01
the area to the right of z0.01 is 0.01 and the area to the left of z0.01 is 1 - 0.01 = 0.99
Looking at the t distribution table,
z = 2.998
Margin of error = 2.998 × 1.12/√8
= 1.19
the lower limit of this confidence interval is
6.738 - 1.19 = 5.548
the upper limit of this confidence interval is
6.738 + 1.19 = 7.928
We have 98 % confidence that the true mean time a student sleeps per night is between 5.548 hours and 7.928 hours.