A pharmaceutical company makes tranquilizers. We wish to find the 89% confidence interval. It is assumed that the distribution for the length of time they last is approximately normal. The standard deviation for all lengths of time for the tranquilizer is 0.21 hours. Researchers in a hospital used the drug on a random sample of 33 patients. The effective period of the tranquilizer for each patient (in hours) is shown in the table below. Round all answers to the nearest hundredth. (a) a 2.2 2.79 2.8 2.64 2.53 2.57 2.31 2.25 2.76 2.61 2.11 2.28 2.22 2.81 2.95 2.1 2.33 2.83 2.71 2.96 2.49 2.72 2.23 2.15 2.81 2.45 2.38 2.63 2.54 2.37 2.76 2.27 2.69 = (b) s = (c) σ= _id=965498 2.52 0.21 ✔ Enter DNE if this is unknown. Enter DNE if this is unknown. ⠀
A pharmaceutical company makes tranquilizers. We wish to find the 89% confidence interval. It is assumed that the distribution for the length of time they last is approximately normal. The standard deviation for all lengths of time for the tranquilizer is 0.21 hours. Researchers in a hospital used the drug on a random sample of 33 patients. The effective period of the tranquilizer for each patient (in hours) is shown in the table below. Round all answers to the nearest hundredth. (a) a 2.2 2.79 2.8 2.64 2.53 2.57 2.31 2.25 2.76 2.61 2.11 2.28 2.22 2.81 2.95 2.1 2.33 2.83 2.71 2.96 2.49 2.72 2.23 2.15 2.81 2.45 2.38 2.63 2.54 2.37 2.76 2.27 2.69 = (b) s = (c) σ= _id=965498 2.52 0.21 ✔ Enter DNE if this is unknown. Enter DNE if this is unknown. ⠀
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Author:Amos Gilat
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Transcribed Image Text:**Calculating the 89% Confidence Interval for Tranquilizer Duration**
A pharmaceutical company produces tranquilizers. To find the 89% confidence interval for the average duration of their effect, it is assumed that the duration follows an approximately normal distribution. The standard deviation for all durations is 0.21 hours. A study conducted by researchers involved examining the effect on 33 patients, and the effective period in hours is listed below. All results should be rounded to the nearest hundredth.
**Data Table:**
The effective period in hours for each patient is presented in a table with seven columns and five rows:
- 2.2, 2.79, 2.8, 2.64, 2.53
- 2.57, 2.31, 2.25, 2.76, 2.61
- 2.11, 2.28, 2.22, 2.8, 2.95
- 2.31, 2.33, 2.71, 2.96
- 2.45, 2.8, 2.25, 2.81
- 2.76, 2.27, 2.69, 2.37
**Calculation steps:**
(a) The sample mean \(\bar{x}\) is calculated to be 2.52.
(b) The standard deviation (σ) is given as 0.21.
(c) The sample size (\( n \)) is 33.
(d) The critical value (z*) for an 89% confidence interval is 1.60.
(e) The confidence interval is calculated, ensuring to fill in the correct values when following the formula \(\bar{x} \pm \left( z* \times \frac{\sigma}{\sqrt{n}} \right)\).
**Conclusion:**
With 89% confidence, the mean time in hours for the tranquilizer's effect is estimated to fall within a specified interval, determined by completing the confidence interval calculation.
Expert Solution
Step 1
Data given
2.2,2.79,2.8,2.64,2.53,2.57,2.31,2.25,2.76,2.61,2.11,2.28,2.22,2.81,2.95,2.1,2.33,2.83,2.71,2.96,2.49,2.72,2.23,2.15,2.81,2.45,2.38,2.63,2.54,2.37,2.76,2.27,2.69
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