Final answer:
To calculate the 90% confidence interval for the mean number of visitors to social networking sites, we use the t-distribution with the given mean, standard deviation, and sample size, and we are 90 percent confident that the true mean lies between 11.058 million and 15.522 million visitors.
Step-by-step explanation:
To find the 90% confidence interval of the true mean number of visitors to the social networking sites, we will use the following formula for confidence interval when the sample size is small (under 30) and the standard deviation is known:
Confidence Interval = µ ± (t-value)(σ/√n)
Where ± represents the plus-minus sign, σ is the sample standard deviation, √n is the square root of the sample size, and t-value comes from the t-distribution table corresponding to our confidence level and degrees of freedom (df = n - 1).
In this case, the sample size (n) is 9, the sample mean (µ) is 13.29 million, and the sample standard deviation (σ) is 3.6 million. The degrees of freedom (df) are 9 - 1 = 8.
Using the t-distribution table for df = 8 and a 90% confidence interval, we find that the t-value is approximately 1.86.
Now let's calculate the margin of error:
Margin of Error = t-value × (σ/√n) = 1.86 × (3.6/√9) = 1.86 × 1.2 = 2.232 million
Then, the confidence interval is:
13.29 million ± 2.232 million
So, we are 90 percent confident that the true mean number of visitors to these social networking sites is between:
11.058 million and 15.522 million