Answer :
We wish to test the null hypothesis
[tex]$$
H_0: \mu = 98.6^\circ F
$$[/tex]
against the alternative hypothesis
[tex]$$
H_a: \mu \neq 98.6^\circ F.
$$[/tex]
The steps for the one-sample [tex]\( t \)[/tex]-test are detailed below.
1. \textbf{Compute the sample mean.}
The sample mean is calculated as
[tex]$$
\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i.
$$[/tex]
For the given sample of 10 temperatures, the computed sample mean is
[tex]$$
\bar{x} \approx 98.12^\circ F.
$$[/tex]
2. \textbf{Compute the sample standard deviation.}
The sample standard deviation (with Bessel's correction) is
[tex]$$
s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1}}.
$$[/tex]
The value obtained is
[tex]$$
s \approx 0.92^\circ F.
$$[/tex]
3. \textbf{Compute the standard error of the mean.}
The standard error (SE) is found by
[tex]$$
\text{SE} = \frac{s}{\sqrt{n}},
$$[/tex]
where [tex]\( n = 10 \)[/tex]. Thus,
[tex]$$
\text{SE} \approx \frac{0.92}{\sqrt{10}} \approx 0.29.
$$[/tex]
4. \textbf{Calculate the [tex]\( t \)[/tex]-statistic.}
The [tex]\( t \)[/tex]-statistic is given by
[tex]$$
t = \frac{\bar{x} - \mu_0}{\text{SE}},
$$[/tex]
where [tex]\( \mu_0 = 98.6^\circ F \)[/tex]. Substituting the values,
[tex]$$
t \approx \frac{98.12 - 98.6}{0.29} \approx -1.65.
$$[/tex]
5. \textbf{Determine the degrees of freedom.}
Since there are 10 observations, the degrees of freedom is
[tex]$$
df = n - 1 = 9.
$$[/tex]
6. \textbf{Compute the two-tailed [tex]\( p \)[/tex]-value.}
For a two-tailed test, the [tex]\( p \)[/tex]-value is calculated as
[tex]$$
p = 2 \, P(T \geq |t|),
$$[/tex]
where [tex]\( T \)[/tex] follows a [tex]\( t \)[/tex]-distribution with 9 degrees of freedom. The computed [tex]\( p \)[/tex]-value is approximately
[tex]$$
p \approx 0.133.
$$[/tex]
7. \textbf{Conclusion.}
At a common significance level such as [tex]\( \alpha = 0.05 \)[/tex], the [tex]\( p \)[/tex]-value of approximately 0.133 is greater than [tex]\( \alpha \)[/tex]. Thus, there is not enough evidence to reject the null hypothesis. We conclude that there is insufficient evidence to claim that the population mean body temperature is different from [tex]\( 98.6^\circ F \)[/tex].
In summary, the test yielded:
- Sample mean: [tex]\( \bar{x} \approx 98.12^\circ F \)[/tex]
- Sample standard deviation: [tex]\( s \approx 0.92^\circ F \)[/tex]
- [tex]\( t \)[/tex]-statistic: [tex]\( t \approx -1.65 \)[/tex]
- Degrees of freedom: 9
- Two-tailed [tex]\( p \)[/tex]-value: [tex]\( p \approx 0.133 \)[/tex]
Since the [tex]\( p \)[/tex]-value is not less than the significance level, we fail to reject the null hypothesis that [tex]\( \mu = 98.6^\circ F \)[/tex].
[tex]$$
H_0: \mu = 98.6^\circ F
$$[/tex]
against the alternative hypothesis
[tex]$$
H_a: \mu \neq 98.6^\circ F.
$$[/tex]
The steps for the one-sample [tex]\( t \)[/tex]-test are detailed below.
1. \textbf{Compute the sample mean.}
The sample mean is calculated as
[tex]$$
\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i.
$$[/tex]
For the given sample of 10 temperatures, the computed sample mean is
[tex]$$
\bar{x} \approx 98.12^\circ F.
$$[/tex]
2. \textbf{Compute the sample standard deviation.}
The sample standard deviation (with Bessel's correction) is
[tex]$$
s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1}}.
$$[/tex]
The value obtained is
[tex]$$
s \approx 0.92^\circ F.
$$[/tex]
3. \textbf{Compute the standard error of the mean.}
The standard error (SE) is found by
[tex]$$
\text{SE} = \frac{s}{\sqrt{n}},
$$[/tex]
where [tex]\( n = 10 \)[/tex]. Thus,
[tex]$$
\text{SE} \approx \frac{0.92}{\sqrt{10}} \approx 0.29.
$$[/tex]
4. \textbf{Calculate the [tex]\( t \)[/tex]-statistic.}
The [tex]\( t \)[/tex]-statistic is given by
[tex]$$
t = \frac{\bar{x} - \mu_0}{\text{SE}},
$$[/tex]
where [tex]\( \mu_0 = 98.6^\circ F \)[/tex]. Substituting the values,
[tex]$$
t \approx \frac{98.12 - 98.6}{0.29} \approx -1.65.
$$[/tex]
5. \textbf{Determine the degrees of freedom.}
Since there are 10 observations, the degrees of freedom is
[tex]$$
df = n - 1 = 9.
$$[/tex]
6. \textbf{Compute the two-tailed [tex]\( p \)[/tex]-value.}
For a two-tailed test, the [tex]\( p \)[/tex]-value is calculated as
[tex]$$
p = 2 \, P(T \geq |t|),
$$[/tex]
where [tex]\( T \)[/tex] follows a [tex]\( t \)[/tex]-distribution with 9 degrees of freedom. The computed [tex]\( p \)[/tex]-value is approximately
[tex]$$
p \approx 0.133.
$$[/tex]
7. \textbf{Conclusion.}
At a common significance level such as [tex]\( \alpha = 0.05 \)[/tex], the [tex]\( p \)[/tex]-value of approximately 0.133 is greater than [tex]\( \alpha \)[/tex]. Thus, there is not enough evidence to reject the null hypothesis. We conclude that there is insufficient evidence to claim that the population mean body temperature is different from [tex]\( 98.6^\circ F \)[/tex].
In summary, the test yielded:
- Sample mean: [tex]\( \bar{x} \approx 98.12^\circ F \)[/tex]
- Sample standard deviation: [tex]\( s \approx 0.92^\circ F \)[/tex]
- [tex]\( t \)[/tex]-statistic: [tex]\( t \approx -1.65 \)[/tex]
- Degrees of freedom: 9
- Two-tailed [tex]\( p \)[/tex]-value: [tex]\( p \approx 0.133 \)[/tex]
Since the [tex]\( p \)[/tex]-value is not less than the significance level, we fail to reject the null hypothesis that [tex]\( \mu = 98.6^\circ F \)[/tex].
