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A data set includes 110 body temperatures of healthy adult humans with a mean of [tex]$98.1^{\circ} F$[/tex] and a standard deviation of [tex]$0.64^{\circ} F$[/tex]. Construct a [tex]$99\%$[/tex] confidence interval estimate of the mean body temperature of all healthy humans. What does the sample suggest about the use of [tex]$98.6^{\circ} F$[/tex] as the mean body temperature?

What is the confidence interval estimate of the population mean [tex]$\mu$[/tex]?
[tex]$\square^{\circ} F \ \textless \ \mu \ \textless \ \square^{\circ} F$[/tex]
(Round to three decimal places as needed.)

What does this suggest about the use of [tex]$98.6^{\circ} F$[/tex] as the mean body temperature?
A. This suggests that the mean body temperature could very possibly be [tex]$98.6^{\circ} F$[/tex].
B. This suggests that the mean body temperature is lower than [tex]$98.6^{\circ} F$[/tex].
C. This suggests that the mean body temperature is higher than [tex]$98.6^{\circ} F$[/tex].

Sagot :

GinnyAnsweridn
To construct a 99% confidence interval estimate for the mean body temperature of all healthy humans based on the given data, follow these steps:

1. Determine the Sample Size ([tex]\(n\)[/tex]):
- Given: [tex]\( n = 110 \)[/tex]

2. Calculate the Sample Mean ([tex]\(\bar{x}\)[/tex]):
- Given: [tex]\( \bar{x} = 98.1^{\circ} F \)[/tex]

3. Calculate the Sample Standard Deviation ([tex]\(s\)[/tex]):
- Given: [tex]\( s = 0.64^{\circ} F \)[/tex]

4. Find the Confidence Level and Corresponding Critical Value:
- Confidence level: 99%
- Degrees of freedom ([tex]\(df\)[/tex]) = [tex]\(n - 1 = 110 - 1 = 109\)[/tex]
- Using the t-distribution table, look up the critical value ([tex]\(t^\)[/tex]) that corresponds to a 99% confidence level with 109 degrees of freedom. The critical value [tex]\(t^\)[/tex] for 99% confidence and 109 degrees of freedom is approximately [tex]\(2.626\)[/tex].

5. Calculate the Margin of Error (ME):
- Formula: [tex]\( ME = t^* \times \frac{s}{\sqrt{n}} \)[/tex]
- [tex]\( ME = 2.626 \times \frac{0.64}{\sqrt{110}} \)[/tex]
- Calculate the margin of error:
[tex]\[ ME \approx 2.626 \times 0.061 \approx 0.160 \][/tex]

6. Construct the Confidence Interval:
- Lower limit: [tex]\(\bar{x} - ME = 98.1 - 0.160 = 97.94\)[/tex]
- Upper limit: [tex]\(\bar{x} + ME = 98.1 + 0.160 = 98.26\)[/tex]

Therefore, the 99% confidence interval estimate of the population mean [tex]\(\mu\)[/tex] is:
[tex]\[ 97.94^{\circ} F < \mu < 98.26^{\circ} F \][/tex]

7. Interpret the Confidence Interval:
- The interval [tex]\(97.94^{\circ} F\)[/tex] to [tex]\(98.26^{\circ} F\)[/tex] does not include [tex]\(98.6^{\circ} F\)[/tex].
- This suggests that the mean body temperature is unlikely to be [tex]\(98.6^{\circ} F\)[/tex]. In fact, the entire confidence interval is below [tex]\(98.6^{\circ} F\)[/tex].

Based on this confidence interval, the usage of [tex]\(98.6^{\circ} F\)[/tex] as the mean body temperature appears questionable. The sample suggests that the mean body temperature is lower than [tex]\(98.6^{\circ} F\)[/tex].

So, the answer to what this suggests is:
B. This suggests that the mean body temperature is lower than [tex]\(98.6^{\circ} F\)[/tex].
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