Answer:
Step-by-step explanation:
The question is incomplete. The complete question is:
A consumer rights advocate wants to determine the lifetime of a certain type of tablet computer. She randomly selects a sample of 29 of these tablets, and calculates the mean and standard deviation to be x = 6.7 years and s = 2.3 years. Find a 95% confidence interval for the true mean lifetime (in years) of this type of tablet.
Solution:
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
s = sample standard deviation = 2.3
n = number of samples = 29
x = 6.7
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 t score
In order to use the t distribution, we would determine the degree of freedom, df for the sample.
df = n - 1 = 29 - 1 = 28
Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05
α/2 = 0.05/2 = 0.025
the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975
Looking at the t distribution table,
t = 2.048
Margin of error = 2.048 × 2.3/√29
= 0.87
the lower limit of this confidence interval is
6.7 - 0.87 = 5.83
the upper limit of this confidence interval is
6.7 + 0.87 = 7.57
Therefore, we are 95% confident that true mean lifetime of a certain type of tablet computer lies between 5.83 years and 7.57 years
