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.

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 (\(\bar{x}\)): 106
- Standard deviation (\(\sigma\)): 11
- Sample size (\(n\)): 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 (\(Z_{\alpha/2}\)) is 1.645.

### Step 3: Calculate the standard error (SE)
The standard error of the mean is given by:
\[
\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{11}{\sqrt{70}} \approx \frac{11}{8.3666} \approx 1.314
\]

### Step 4: Compute the margin of error (ME)
The margin of error is given by:
\[
\text{ME} = Z_{\alpha/2} \times \text{SE} = 1.645 \times 1.314 \approx 2.161
\]

### Step 5: Determine the confidence interval
The confidence interval is calculated as:
\[
\text{Confidence Interval} = \bar{x} \pm \text{ME} = 106 \pm 2.161
\]
This gives us:
\[
\text{Lower limit} = 106 - 2.161 \approx 103.8
\]
\[
\text{Upper limit} = 106 + 2.161 \approx 108.2
\]

### 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 \([103.8, 108.2]\).

Would you like any further details or have any questions?

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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.

Learn how to calculate a 90% confidence interval for the mean IQ score using a sample mean of 106, standard deviation of 11, and a sample size of 70. This problem involves statistical concepts such as confidence intervals, standard error, and the Z-score. Suitable for advanced high school or college-level students.

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