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Answer :
We start by noting that we are testing the null hypothesis
[tex]$$
H_0: \mu = 98.6^\circ F
$$[/tex]
against the two-sided alternative
[tex]$$
H_1: \mu \neq 98.6^\circ F.
$$[/tex]
The sample data of 10 body temperatures (in °F) is:
[tex]$$
98.5,\, 98.2,\, 99.0,\, 96.3,\, 98.3,\, 98.7,\, 97.2,\, 99.1,\, 98.7,\, 97.2.
$$[/tex]
Step 1. Calculate the Sample Mean
The sample mean is computed by summing all observations and dividing by the sample size ([tex]$n=10$[/tex]):
[tex]$$
\bar{x} = 98.12^\circ F.
$$[/tex]
Step 2. Calculate the Sample Standard Deviation
The sample standard deviation, calculated with Bessel's correction (i.e. dividing by [tex]$n-1$[/tex]), is found to be
[tex]$$
s \approx 0.9187^\circ F.
$$[/tex]
Step 3. Compute the Standard Error of the Mean
The standard error (SE) is given by
[tex]$$
\text{SE} = \frac{s}{\sqrt{n}} = \frac{0.9187}{\sqrt{10}}.
$$[/tex]
Step 4. Calculate the Test Statistic
The [tex]$t$[/tex]-statistic for testing the mean is obtained via
[tex]$$
t = \frac{\bar{x} - \mu_0}{\text{SE}} = \frac{98.12 - 98.6}{0.9187 / \sqrt{10}},
$$[/tex]
which evaluates to
[tex]$$
t \approx -1.6522.
$$[/tex]
Step 5. Determine the p-value
With [tex]$n=10$[/tex], there are
[tex]$$
\text{df} = 10 - 1 = 9
$$[/tex]
degrees of freedom. Since this is a two-tailed test, we calculate the p-value corresponding to [tex]$t=-1.6522$[/tex]. This p-value is found to be
[tex]$$
\text{p-value} \approx 0.1329.
$$[/tex]
Step 6. State the Conclusion
Since the p-value ([tex]$\approx 0.1329$[/tex]) is larger than the common significance level (say, [tex]$\alpha = 0.05$[/tex]), we do not reject the null hypothesis. In other words, there is not enough evidence to conclude that the actual mean body temperature is different from [tex]$98.6^\circ F$[/tex].
Final Answer:
The sample has a mean of [tex]$98.12^\circ F$[/tex] and a standard deviation of approximately [tex]$0.9187^\circ F$[/tex]. The computed [tex]$t$[/tex]-statistic is [tex]$-1.6522$[/tex], with an associated two-tailed p-value of approximately [tex]$0.1329$[/tex]. Therefore, we do not reject [tex]$H_0$[/tex] at the [tex]$\alpha=0.05$[/tex] level.
[tex]$$
H_0: \mu = 98.6^\circ F
$$[/tex]
against the two-sided alternative
[tex]$$
H_1: \mu \neq 98.6^\circ F.
$$[/tex]
The sample data of 10 body temperatures (in °F) is:
[tex]$$
98.5,\, 98.2,\, 99.0,\, 96.3,\, 98.3,\, 98.7,\, 97.2,\, 99.1,\, 98.7,\, 97.2.
$$[/tex]
Step 1. Calculate the Sample Mean
The sample mean is computed by summing all observations and dividing by the sample size ([tex]$n=10$[/tex]):
[tex]$$
\bar{x} = 98.12^\circ F.
$$[/tex]
Step 2. Calculate the Sample Standard Deviation
The sample standard deviation, calculated with Bessel's correction (i.e. dividing by [tex]$n-1$[/tex]), is found to be
[tex]$$
s \approx 0.9187^\circ F.
$$[/tex]
Step 3. Compute the Standard Error of the Mean
The standard error (SE) is given by
[tex]$$
\text{SE} = \frac{s}{\sqrt{n}} = \frac{0.9187}{\sqrt{10}}.
$$[/tex]
Step 4. Calculate the Test Statistic
The [tex]$t$[/tex]-statistic for testing the mean is obtained via
[tex]$$
t = \frac{\bar{x} - \mu_0}{\text{SE}} = \frac{98.12 - 98.6}{0.9187 / \sqrt{10}},
$$[/tex]
which evaluates to
[tex]$$
t \approx -1.6522.
$$[/tex]
Step 5. Determine the p-value
With [tex]$n=10$[/tex], there are
[tex]$$
\text{df} = 10 - 1 = 9
$$[/tex]
degrees of freedom. Since this is a two-tailed test, we calculate the p-value corresponding to [tex]$t=-1.6522$[/tex]. This p-value is found to be
[tex]$$
\text{p-value} \approx 0.1329.
$$[/tex]
Step 6. State the Conclusion
Since the p-value ([tex]$\approx 0.1329$[/tex]) is larger than the common significance level (say, [tex]$\alpha = 0.05$[/tex]), we do not reject the null hypothesis. In other words, there is not enough evidence to conclude that the actual mean body temperature is different from [tex]$98.6^\circ F$[/tex].
Final Answer:
The sample has a mean of [tex]$98.12^\circ F$[/tex] and a standard deviation of approximately [tex]$0.9187^\circ F$[/tex]. The computed [tex]$t$[/tex]-statistic is [tex]$-1.6522$[/tex], with an associated two-tailed p-value of approximately [tex]$0.1329$[/tex]. Therefore, we do not reject [tex]$H_0$[/tex] at the [tex]$\alpha=0.05$[/tex] level.
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