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Jeremy is conducting a survey about his coworkers' in-office water consumption to encourage management to install more water dispensers in their location. He found that the population mean is 112.5 ounces with a standard deviation of 37.5. Jeremy has a sample size of 96.

Complete the equation that Jeremy can use to find the interval in which he can be 95% sure that the sample mean will lie:

112.5, 37.5, 96, 98, 75, 150, 3



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GinnyAnswer
Jeremy is conducting a survey about his coworkers' in-office water consumption to encourage management to install more water dispensers at their location. He found that the population mean is 112.5 ounces with a standard deviation of 37.5. Jeremy has a sample size of 96.

Complete the equation that Jeremy can use to find the interval in which he can be 95% sure that the sample mean will lie.

1. Identify given values and required equation:
- Population mean (μ): 112.5
- Population standard deviation (σ): 37.5
- Sample size (n): 96
- Confidence level: 95% (which corresponds to a z-score of 1.96 for a two-tailed test)

2. Calculate the standard error (SE):
[tex]\[ SE = \frac{\sigma}{\sqrt{n}} \][/tex]
Plugging in the values:
[tex]\[ SE = \frac{37.5}{\sqrt{96}} = 3.83 \][/tex]

3. Calculate the margin of error (ME):
[tex]\[ ME = z \times SE \][/tex]
Using the z-score for a 95% confidence interval which is 1.96:
[tex]\[ ME = 1.96 \times 3.83 = 7.5068 \][/tex]

4. Determine the confidence interval (CI):
- Lower bound:
[tex]\[ \text{Lower bound} = \mu - ME = 112.5 - 7.5068 = 104.99 \][/tex]
- Upper bound:
[tex]\[ \text{Upper bound} = \mu + ME = 112.5 + 7.5068 = 120.01 \][/tex]

Therefore, Jeremy can be 95% confident that the sample mean will lie between 104.99 and 120.01 ounces. The values used in the calculation include 96 (sample size), 1125/10 (which results from the mean scaled from 112.5 for convenience), 75 (SE calculation intermediary result), 37.5 (standard deviation), 96 (sample size again), and 112.5 (population mean).

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