A toy manufacturer wants to see how long, on average, a new toy captures children’s attention. He tests 17 children selected at random and finds that their
mean attention span is 31 minutes with a standard deviation of 6 minutes. If we assume that attention spans are normally distributed, find a confidence
interval for the mean attention span of children playing with this new toy. Give the lower limit and upper limit of the confidence interval.
Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)
Lower limit:
Upper limit:

Question

A toy manufacturer wants to see how long, on average, a new toy captures children’s attention. He tests 17 children selected at random and finds that their
mean attention span is 31 minutes with a standard deviation of 6 minutes. If we assume that attention spans are normally distributed, find a

confidence
interval for the mean attention span of children playing with this new toy. Give the lower limit and upper limit of the

confidence interval.
Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)
Lower limit:
Upper limit:

UpStudy AI · Accepted Answer

**Step 1. Identify the given values**

- Sample mean: $$ \bar{x} = 31 $$
- Sample standard deviation: $$ s = 6 $$
- Sample size: $$ n = 17 $$
- Confidence level: $$ 90\% $$

**Step 2. Since the population standard deviation is unknown and the data is normally distributed, we use the t-distribution.**  
The degrees of freedom (df) is:
$$
\text{df} = n - 1 = 17 - 1 = 16
$$

**Step 3. Find the t-critical value $$ t_{\alpha/2} $$ for a $$ 90\% $$ confidence level.**  
For a $$ 90\% $$ confidence interval, $$ \alpha = 1 - 0.90 = 0.10 $$, and $$ \alpha/2 = 0.05 $$.  
Using a t-distribution table or calculator for $$ \text{df} = 16 $$, we get:
$$
t_{0.05,16} \approx 1.746
$$

**Step 4. Compute the standard error (SE) of the mean.**  
$$
\text{SE} = \frac{s}{\sqrt{n}} = \frac{6}{\sqrt{17}}
$$
$$
\sqrt{17} \approx 4.123
$$
$$
\text{SE} \approx \frac{6}{4.123} \approx 1.455
$$

**Step 5. Calculate the margin of error (ME).**  
$$
\text{ME} = t_{\alpha/2} \times \text{SE} \approx 1.746 \times 1.455 \approx 2.541
$$

**Step 6. Determine the confidence interval.**  
The $$ 90\% $$ confidence interval for the mean is given by:
$$
\bar{x} \pm \text{ME} = 31 \pm 2.541
$$
Calculating the limits:
- Lower limit: 
$$
31 - 2.541 \approx 28.5
$$
- Upper limit:
$$
31 + 2.541 \approx 33.5
$$

**Final Answer:**  
Lower limit: $$ 28.5 $$ minutes  
Upper limit: $$ 33.5 $$ minutes
Lower limit: 28.5 minutes Upper limit: 33.5 minutes

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