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Answer :
Using the t-distribution, as we have the standard deviation for the sample, it is found that the 99% confidence interval for the true mean number of visitors on summer days is given by:
c. 92.664 to 103.940 visitors
What is a t-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm t\frac{s}{\sqrt{n}}[/tex]
In which:
- [tex]\overline{x}[/tex] is the sample mean.
- t is the critical value.
- n is the sample size.
- s is the standard deviation for the sample.
The critical value, using a t-distribution calculator, for a two-tailed 99% confidence interval, with 30 - 1 = 29 df, is t = 2.756.
The other parameters are given as follows:
[tex]\overline{x} = 98.3, s = 11.2, n = 30[/tex].
Hence, the bounds of the interval are given by:
[tex]\overline{x} - t\frac{s}{\sqrt{n}} = 98.3 - 2.756\frac{11.2}{\sqrt{30}} = 92.664[/tex]
[tex]\overline{x} + t\frac{s}{\sqrt{n}} = 98.3 + 2.756\frac{11.2}{\sqrt{30}} = 103.94[/tex]
Hence option c is correct.
More can be learned about the t-distribution at https://brainly.com/question/16162795
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Option c. is correct. The 99% confidence interval for the true mean number of visitors on summer days is approximately 92.68 to 103.92 visitors. This calculation uses the t-distribution and the sample's standard deviation to determine the interval.
To calculate the 99% confidence interval for the true mean number of visitors on summer days, given a sample mean (
X b..ar) of 98.3 and a sample standard deviation (Sx) of 11.2 with n=30, we'll use the t-distribution because we are working with a sample size less than 30 and do not know the population standard deviation.
First, we determine the t-score for a 99% confidence interval with 29 degrees of freedom (n-1). Using a t-distribution table or calculator, we find the t-score that corresponds to the 99% confidence level. We then calculate the margin of error (ME) using the formula ME = t-score * (Sx/√n).
Margin of Error (ME) calculation:
ME = t-score * (11.2/√30)
Plugging in the values:
ME = (around 2.75) * (11.2/√30)
ME ≈ 2.75 * 2.045 = 5.62
Now, calculate the confidence interval using the formula:
lower limit =Xb..ar - ME
upper limit =Xb..ar + ME
Plugging in our sample mean and margin of error:
lower limit = 98.3 - 5.62 = 92.68
upper limit = 98.3 + 5.62 = 103.92
Therefore, the 99% confidence interval for the true mean number of visitors on summer days is approximately 92.68 to 103.92 visitors. Correct option according to the given choices would be (c) 92.664 to 103.940 visitors.
