Zc 9070=1.645 Z, 95%0=1696 Zc 99%0=2.575 8. # 6.2.1 (on page 152 of 11 th edition) of your textbook. For this problem, find the 90%, 95%, and 99% confidence intervals for the mean number of heartbeats per minute for a certain population! n =49 J=1o x=90 文士 Ze メさて。。 ×さ Z。%。 V49 149 CI. 187.65,92.35 C.I, 9915 /86.2,93.48
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![### Confidence Intervals for Heartbeat Rates
#### Problem Statement:
We wish to estimate the average number of heartbeats per minute for a certain population. The average number of heartbeats per minute for a sample of 49 subjects was found to be 90. Assume that these 49 patients constitute a random sample, and that the population is normally distributed with a standard deviation of 10.
#### Task:
Find the 90%, 95%, and 99% confidence intervals for the mean number of heartbeats per minute for a certain population.
#### Data Provided:
- Sample size (\( n \)) = 49
- Standard deviation (\( \sigma \)) = 10
- Sample mean (\( \bar{X} \)) = 90
- Critical values for Z at different confidence levels:
- \( Z_{c \, 90\%} = 1.645 \)
- \( Z_{c \, 95\%} = 1.96 \)
- \( Z_{c \, 99\%} = 2.575 \)
#### Formulas:
The formula to calculate the confidence interval is given by:
\[ \bar{X} \pm Z_{c} \left( \frac{\sigma}{\sqrt{n}} \right) \]
#### Calculations:
1. **90% Confidence Interval:**
\[
90 \pm 1.645 \left( \frac{10}{\sqrt{49}} \right)
\]
Simplified:
\[
90 \pm 1.645 \left( \frac{10}{7} \right)
\]
\[
90 \pm 1.645 \times 1.4286 \approx 90 \pm 2.35
\]
**90% Confidence Interval:** \( (87.65, 92.35) \)
2. **95% Confidence Interval:**
\[
90 \pm 1.96 \left( \frac{10}{\sqrt{49}} \right)
\]
Simplified:
\[
90 \pm 1.96 \left( \frac{10}{7} \right)
\]
\[
90 \pm 1.96 \times 1.4286 \approx 90 \pm 2.8](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1bc805eb-60ed-457b-ab8e-02c516698c74%2Fb908a34f-1b61-4cac-b6a1-7be523f118b2%2Fx84rs5_processed.jpeg&w=3840&q=75)
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