Construct a 90% confidence interval using the inequality d-t sd+t To test the effectiveness of a new drug that is reported to increase the number of hours of sleep patients get during the night, researchers randomly select 13 patients and record the number of hours of sleep each gets with and without the drug Patient Sleop without the drug 26 45 Sleep using the drug 51 5.4 7.1 52 34 48 4.6 65 52 57 54 3.1 53 10 11 12 13 D 44 35 39 3.8 22 35 Calculate d for each patient by subtracting the number of hours of sleep with the drug from the number without the drug. The confidence interval is< (Type an integer or a decimal Round to two decimal places as needed.)

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**Construct a 90% Confidence Interval**

To test the effectiveness of a new drug that is reported to increase the number of hours of sleep patients get during the night, researchers randomly select 13 patients and record the number of hours of sleep each gets with and without the new drug.

### Data Table

| Patient | 1   | 2   | 3   | 4   | 5   | 6   | 7   | 8   | 9   | 10  | 11  | 12  | 13  |
|---------|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|
| Sleep without the drug | 2.6 | 5.4 | 4.2 | 3.8 | 3.9 | 4.4 | 5.5 | 6.7 | 2.8 | 3.9 | 3.2 | 4.5 | 4.9 |
| Sleep using the drug   | 5.3 | 6.4 | 6.9 | 4.4 | 4.8 | 5.4 | 6.6 | 7.5 | 5.3 | 5.7 | 5.5 | 5.3 | 6.1 |

### Instructions

Calculate \( d \) for each patient by subtracting the number of hours of sleep without the drug from the number with the drug. The confidence interval is \( \bar{d} - t \frac{s_d}{\sqrt{n}} < \mu_d < \bar{d} + t \frac{s_d}{\sqrt{n}} \).

**Note:** Type an integer or a decimal and round to two decimal places as needed.

Enter your answer in each of the answer boxes.

---

This page allows students to learn about constructing confidence intervals and gain hands-on experience by working with real data.
Transcribed Image Text:Here's a transcription suitable for an educational website: --- **Construct a 90% Confidence Interval** To test the effectiveness of a new drug that is reported to increase the number of hours of sleep patients get during the night, researchers randomly select 13 patients and record the number of hours of sleep each gets with and without the new drug. ### Data Table | Patient | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | |---------|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----|-----| | Sleep without the drug | 2.6 | 5.4 | 4.2 | 3.8 | 3.9 | 4.4 | 5.5 | 6.7 | 2.8 | 3.9 | 3.2 | 4.5 | 4.9 | | Sleep using the drug | 5.3 | 6.4 | 6.9 | 4.4 | 4.8 | 5.4 | 6.6 | 7.5 | 5.3 | 5.7 | 5.5 | 5.3 | 6.1 | ### Instructions Calculate \( d \) for each patient by subtracting the number of hours of sleep without the drug from the number with the drug. The confidence interval is \( \bar{d} - t \frac{s_d}{\sqrt{n}} < \mu_d < \bar{d} + t \frac{s_d}{\sqrt{n}} \). **Note:** Type an integer or a decimal and round to two decimal places as needed. Enter your answer in each of the answer boxes. --- This page allows students to learn about constructing confidence intervals and gain hands-on experience by working with real data.
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