Solved on Nov 07, 2023

Construct a 99% confidence interval for the mean body temperature of 105 healthy adults with $\bar{x}=98.7^{\circ}F$ and $s=0.64^{\circ}F$ . Does the interval suggest the use of $98.6^{\circ}F$ as the mean body temperature?
A. The interval suggests the mean could be $98.6^{\circ}F$ . B. The interval suggests the mean is higher than $98.6^{\circ}F$ . C. The interval suggests the mean is lower than $98.6^{\circ}F$ .

STEP 1

Assumptions1. The sample size is105. The sample mean is $98.7^{\circ} \mathrm{}$
3. The standard deviation is $0.64^{\circ} \mathrm{}$
4. We are constructing a99% confidence interval5. We are using a t-distribution due to the sample size being less than30

STEP 2

First, we need to find the t-value for a99% confidence interval with104 degrees of freedom (n-1). We can look this up in a t-distribution table.
$t = t_{0.005,104}$

STEP 3

The t-value for a99% confidence interval with104 degrees of freedom is approximately2.626.
$t =2.626$

STEP 4

Now, we can calculate the margin of error for the confidence interval. The formula for the margin of error is $= t \times \frac{s}{\sqrt{n}}$

STEP 5

Plug in the values for t, s (standard deviation), and n (sample size) to calculate the margin of error.
$=2.626 \times \frac{0.64}{\sqrt{105}}$

STEP 6

Calculate the margin of error.
$\approx0.164^{\circ} \mathrm{}$

STEP 7

Now that we have the margin of error, we can construct the99% confidence interval. The formula for the confidence interval is $\bar{x} - < \mu < \bar{x} +$

STEP 8

Plug in the values for $\bar{x}$ (sample mean) and (margin of error) to calculate the confidence interval.
$98.7^{\circ} \mathrm{} -0.164^{\circ} \mathrm{} < \mu <98.7^{\circ} \mathrm{} +0.164^{\circ} \mathrm{}$

STEP 9

Calculate the confidence interval.
$98.536^{\circ} \mathrm{} < \mu <98.864^{\circ} \mathrm{}$

STEP 10

The mean body temperature of $98.6^{\circ} \mathrm{}$ falls within the99% confidence interval. This suggests that the mean body temperature could very possibly be $98.6^{\circ} \mathrm{}$ .
The correct answer is A. This suggests that the mean body temperature could very possibly be $98.6^{\circ} \mathrm{}$ .

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STEP 1

Assumptions1. The sample size is105. The sample mean is $98.7^{\circ} \mathrm{}$
3. The standard deviation is $0.64^{\circ} \mathrm{}$
4. We are constructing a99% confidence interval5. We are using a t-distribution due to the sample size being less than30

STEP 2

First, we need to find the t-value for a99% confidence interval with104 degrees of freedom (n-1). We can look this up in a t-distribution table.
$t = t_{0.005,104}$

STEP 3

The t-value for a99% confidence interval with104 degrees of freedom is approximately2.626.
$t =2.626$

STEP 4

Now, we can calculate the margin of error for the confidence interval. The formula for the margin of error is $= t \times \frac{s}{\sqrt{n}}$

STEP 5

Plug in the values for t, s (standard deviation), and n (sample size) to calculate the margin of error.
$=2.626 \times \frac{0.64}{\sqrt{105}}$

STEP 6

Calculate the margin of error.
$\approx0.164^{\circ} \mathrm{}$

STEP 7

Now that we have the margin of error, we can construct the99% confidence interval. The formula for the confidence interval is $\bar{x} - < \mu < \bar{x} +$

STEP 8

Plug in the values for $\bar{x}$ (sample mean) and (margin of error) to calculate the confidence interval.
$98.7^{\circ} \mathrm{} -0.164^{\circ} \mathrm{} < \mu <98.7^{\circ} \mathrm{} +0.164^{\circ} \mathrm{}$

STEP 9

Calculate the confidence interval.
$98.536^{\circ} \mathrm{} < \mu <98.864^{\circ} \mathrm{}$

STEP 10

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STEP 1

Assumptions1. The sample size is105. The sample mean is 98.7∘98.7^{\circ} \mathrm{}98.7∘3. The standard deviation is 0.64∘0.64^{\circ} \mathrm{}0.64∘4. We are constructing a99% confidence interval5. We are using a t-distribution due to the sample size being less than30

STEP 2

First, we need to find the t-value for a99% confidence interval with104 degrees of freedom (n-1). We can look this up in a t-distribution table.t=t0.005,104t = t_{0.005,104}t=t0.005,104​

STEP 3

The t-value for a99% confidence interval with104 degrees of freedom is approximately2.626.t=2.626t =2.626t=2.626

STEP 4

Now, we can calculate the margin of error for the confidence interval. The formula for the margin of error is=t×sn = t \times \frac{s}{\sqrt{n}}=t×n​s​

STEP 5

Plug in the values for t, s (standard deviation), and n (sample size) to calculate the margin of error.=2.626×0.64105 =2.626 \times \frac{0.64}{\sqrt{105}}=2.626×105​0.64​

STEP 6

Calculate the margin of error.≈0.164∘ \approx0.164^{\circ} \mathrm{}≈0.164∘

STEP 7

Now that we have the margin of error, we can construct the99% confidence interval. The formula for the confidence interval isxˉ−<μ<xˉ+\bar{x} - < \mu < \bar{x} +xˉ−<μ<xˉ+

STEP 8

Plug in the values for xˉ\bar{x}xˉ (sample mean) and (margin of error) to calculate the confidence interval.98.7∘−0.164∘<μ<98.7∘+0.164∘98.7^{\circ} \mathrm{} -0.164^{\circ} \mathrm{} < \mu <98.7^{\circ} \mathrm{} +0.164^{\circ} \mathrm{}98.7∘−0.164∘<μ<98.7∘+0.164∘

STEP 9

Calculate the confidence interval.98.536∘<μ<98.864∘98.536^{\circ} \mathrm{} < \mu <98.864^{\circ} \mathrm{}98.536∘<μ<98.864∘

STEP 10

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Assumptions1. The sample size is105. The sample mean is $98.7^{\circ} \mathrm{}$
3. The standard deviation is $0.64^{\circ} \mathrm{}$
4. We are constructing a99% confidence interval5. We are using a t-distribution due to the sample size being less than30

First, we need to find the t-value for a99% confidence interval with104 degrees of freedom (n-1). We can look this up in a t-distribution table.
$t = t_{0.005,104}$

The t-value for a99% confidence interval with104 degrees of freedom is approximately2.626.
$t =2.626$

Now, we can calculate the margin of error for the confidence interval. The formula for the margin of error is $= t \times \frac{s}{\sqrt{n}}$

Plug in the values for t, s (standard deviation), and n (sample size) to calculate the margin of error.
$=2.626 \times \frac{0.64}{\sqrt{105}}$

Calculate the margin of error.
$\approx0.164^{\circ} \mathrm{}$

Now that we have the margin of error, we can construct the99% confidence interval. The formula for the confidence interval is $\bar{x} - < \mu < \bar{x} +$

Plug in the values for $\bar{x}$ (sample mean) and (margin of error) to calculate the confidence interval.
$98.7^{\circ} \mathrm{} -0.164^{\circ} \mathrm{} < \mu <98.7^{\circ} \mathrm{} +0.164^{\circ} \mathrm{}$

Calculate the confidence interval.
$98.536^{\circ} \mathrm{} < \mu <98.864^{\circ} \mathrm{}$