Question

he marketing department at an online site wants to identify the average age of its users. a sample of 56 uses is selected. the average age in the sample was 22.5 years with a standard deviation of 5 years. assume the population of consumer ages is normally distributed. construct a 90% confidence interval for the average age of all the users of this site.

Answer 1

The range for the average age of all site visitors is found to be between 21.15 and 23.55.

Define the term confidence interval?

• The likelihood that a quantity will fall between two values near the mean is shown by a confidence interval.
• Confidence intervals quantify how certain or uncertain a sampling technique is.
• They are also utilized in regression analysis and hypothesis testing.

For the given question;

Confidence interval  CI = 90% = 0.9

Sample size n = 56 users

Mean (x) = 22.5 years years

Standard deviation (s) = 5 years

Then formula for the CI is given as,

CI = x + z.s/√n

The steps included are-

1. Subtract 1 from sample size to get the degree of freedom df:

df = 56 - 1 = 55

2. To get teh alpha level.

α = (1 - 0.9)/2 = 0.05

3: check df and α in t-distribution to get z score.

z = 2.003

4: To get sample size,

S = s/√n

S = 5/√56

S = 0.66

z x S = 0.66 x 2.033 = 1.35

5: To get the lower range subtract 1.35 from sample mean

22.5  -  1.35 = 21.15

6:n  To get the upper range, add t 1.35 from sample mean.

22.5 + 1.35 = 23.55

Thus, the confidence interval is 21.15 and 23.55.

To know more about the confidence interval, here

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