Respuesta :
Answer:
The 95% confidence interval = (98.225, 98.655)
Step-by-step explanation:
The formula for Confidence Interval =
Mean ± z × standard deviation/√n
Where Mean = 98.44°F
Standard deviation = 0.30°F
n = number of samples is 10
Because our number if samples is small, we use the t score confidence interval formula
Mean ± t × standard deviation/√n
Degrees of Freedom = n - 1
= 10 - 1 = 9
We find the t score of 95% confidence interval and degrees of freedom 9
= 2.262
Hence,
Confidence interval = 98.44 ± 2.262 × 0.30/√10
= 98.44 ± 0.214592162
Confidence Interval
98.44 - 0.214592162
= 98.225407838
≈ 98.225
98.44 + 0.214592162
= 98.654592162
≈ 98.655
The 95% confidence interval = (98.225, 98.655)
The 95% confidence interval for the mean of all body temperatures is [98.2255, 98.6545] (in degree Fahrenheit)
How to calculate confidence interval for population mean from small sample?
If the sample size is given to be n < 30, then for finding the confidence interval for mean of population from this small sample, we use t-statistic.
- Let the sample mean given as [tex]\overline{x}[/tex] and
- The sample standard deviation s, and
- The sample size = n, and
- The level of significance = [tex]\alpha[/tex]
Then, we get the confidence interval in between the limits
[tex]\overline{x} \pm t_{\alpha/2}\times \dfrac{s}{\sqrt{n}}[/tex]
where [tex]t_{\alpha/2}[/tex] is the critical value of 't' that can be found online or from tabulated values of critical value for specific level of significance and degree of freedom = n - 1
For this case, we have:
- The sample mean given as [tex]\overline{x}[/tex] = 98.44° F
- The sample standard deviation s = 0.30° F, and
- The sample size = n = 10, and
- The level of significance = [tex]\alpha[/tex] = 100% - confidence level = 100% - 95% = 5% = 0.05
The degree of freedom(d.f.) is n-1 = 10-1=9
At d.f. 9, at level of significance 0.05, the critical value of t is [tex]t_{\alpha/2} = 2.262[/tex]
Thus, the limits of the confidence interval is calculated as;
[tex]\overline{x} \pm t_{\alpha/2}\times \dfrac{s}{\sqrt{n}} = 98.44 \pm 2.262 \times \dfrac{0.3}{\sqrt{10}} = 98.44 \pm 0.2145[/tex]
This shows that the confidence interval is [98.44 - 0.2145, 98.44 + 0.2145] which is [98.2255, 98.6545] (in degree Fahrenheit)
Thus, the 95% confidence interval for the mean of all body temperatures is [98.2255, 98.6545] (in degree Fahrenheit)
Learn more about confidence interval for student's t statistic here:
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