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A travel agency call wants to know the average number of calls received per day by its call centre. A random sample of 21 days is selected and the sample mean number of calls received was found to be 166.2 calls with a sample standard deviation of 22.8 calls. Assume that calls received daily are normally distributed. (a) Find a 95% confidence interval for the mean number of daily calls received by the call centre

User Mianos
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Final answer:

The 95% confidence interval for the mean number of daily calls received by the call center, based on the sample data, is approximately (155.69, 176.71).

Step-by-step explanation:

To find the 95% confidence interval for the mean number of daily calls received by the call center, we will use the formula for a confidence interval for a population mean when the population standard deviation is unknown:

Confidence Interval = ºr{x} ± (t* × (s / √n))

  • ºr{x} is the sample mean.
  • t* is the t-score that corresponds to the desired level of confidence and degrees of freedom.
  • s is the sample standard deviation.
  • n is the sample size.

Given:
ºr{x} = 166.2,
s = 22.8, and
n = 21.

Since we do not know the population standard deviation, we will use the t-distribution. For a 95% confidence interval and 20 degrees of freedom (n-1 = 21-1), the t-score is approximately 2.086 (from the t-distribution table or a calculator).

Calculating the margin of error:

Margin of Error = t* × (s / √n) = 2.086 × (22.8 / √21) ≈ 10.51

Therefore, the 95% confidence interval for the mean number of daily calls is:

166.2 ± 10.51 = (155.69, 176.71)

The travel agency can be 95% confident that the mean number of calls lies within this interval.

User Cgaldiolo
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