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

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

d

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