Math Problem Statement

A psychologist is interested in the mean IQ score of a given group of children. It is known that the IQ scores of the group have a standard deviation of 11. The psychologist randomly selects 70 children from this group and finds that their mean IQ score is 106. Based on this sample, find 90% confidence interval for the true mean IQ Score for all children of this group. Then give its lower limit and upper limit. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.

Solution

To find the 90% confidence interval for the true mean IQ score of all children in the group, we'll follow these steps:

Step 1: Identify the given information

  • Sample mean (xˉ\bar{x}): 106
  • Standard deviation (σ\sigma): 11
  • Sample size (nn): 70
  • Confidence level: 90%

Step 2: Find the Z-score for a 90% confidence level

The Z-score corresponding to a 90% confidence level can be found using a Z-table or standard normal distribution calculator. For a 90% confidence level, the Z-score (Zα/2Z_{\alpha/2}) is 1.645.

Step 3: Calculate the standard error (SE)

The standard error of the mean is given by: SE=σn=1170118.36661.314\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{11}{\sqrt{70}} \approx \frac{11}{8.3666} \approx 1.314

Step 4: Compute the margin of error (ME)

The margin of error is given by: ME=Zα/2×SE=1.645×1.3142.161\text{ME} = Z_{\alpha/2} \times \text{SE} = 1.645 \times 1.314 \approx 2.161

Step 5: Determine the confidence interval

The confidence interval is calculated as: Confidence Interval=xˉ±ME=106±2.161\text{Confidence Interval} = \bar{x} \pm \text{ME} = 106 \pm 2.161 This gives us: Lower limit=1062.161103.8\text{Lower limit} = 106 - 2.161 \approx 103.8 Upper limit=106+2.161108.2\text{Upper limit} = 106 + 2.161 \approx 108.2

Final Answer:

  • Lower limit: 103.8
  • Upper limit: 108.2

This means the 90% confidence interval for the true mean IQ score of all children in this group is [103.8,108.2][103.8, 108.2].

Would you like any further details or have any questions?


Here are some related questions for further exploration:

  1. How would the confidence interval change if the confidence level was 95%?
  2. What impact would increasing the sample size have on the confidence interval?
  3. What is the effect of a larger standard deviation on the confidence interval?
  4. How do you interpret the confidence interval in practical terms?
  5. What assumptions are necessary for the validity of the confidence interval in this context?
  6. How would a different sample mean affect the confidence interval?
  7. What happens if the sample size is below 30? Should we still use the Z-distribution?
  8. Can we use the same method if the data is not normally distributed?

Tip: Remember that a larger sample size generally leads to a narrower confidence interval, providing a more precise estimate of the population mean.

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Math Problem Analysis

Mathematical Concepts

Statistics
Confidence Intervals
Normal Distribution

Formulas

Standard Error (SE)
Margin of Error (ME)
Z-score

Theorems

Central Limit Theorem

Suitable Grade Level

Advanced High School or College