Phonics is an instructional method in which children are taught to connect sounds with letters or groups of letters. A sample of 136 first-graders who were learning English were asked to identify as many letter sounds as possible in a period of one minute. The average number of letter sounds identified was 34.1 with a standard deviation of 23.32

Question

(a) Construct a 99.8% confidence interval for the mean number of letter sounds identified in one minute. Round the answers to at least two decimal places..

Answer 1

To construct a 99.8% confidence interval for the mean number of letter sounds identified, we can use the following formula for the confidence interval:

\[ CI = \bar{x} \pm z \left( \frac{s}{\sqrt{n}} \right) \]

Where:

• \(\bar{x}\) is the sample mean
• \(z\) is the z-value that corresponds to the desired confidence level
• \(s\) is the sample standard deviation
• \(n\) is the sample size

Given the values:

• Sample size, \(n = 136\)
• Sample mean, \(\bar{x} = 34.1\)
• Sample standard deviation, \(s = 23.32\)

First, we need to find the z-value for a 99.8% confidence level. We can find this using a z-table or calculator. The critical value for a 99.8% confidence level is approximately \(z \approx 2.666\).

Next, we can calculate the standard error (SE):

\[ SE = \frac{s}{\sqrt{n}} = \frac{23.32}{\sqrt{136}} \approx \frac{23.32}{11.662} \approx 2.00 \]

Now we can calculate the margin of error (ME):

\[ ME = z \cdot SE = 2.666 \cdot 2.00 \approx 5.332 \]

Finally, we can construct the confidence interval:

\[ CI = \bar{x} \pm ME = 34.1 \pm 5.332 \]

Calculating the lower and upper limits:

\[ \text{Lower limit} = 34.1 - 5.332 \approx 28.77 \] \[ \text{Upper limit} = 34.1 + 5.332 \approx 39.43 \]

Thus, the 99.8% confidence interval for the mean number of letter sounds identified in one minute is approximately:

\[ \boxed{(28.77, 39.43)} \]

Answer 2

Incorrect. its 27.80 < u < 40.40

Now if an 80% confidence interval were constructed with these data, would it be wider or narrower than the interval constructed in Part (a)?

Answer 3

I apologize for the earlier miscalculation of the confidence interval. Let's clarify how the width of confidence intervals varies with different confidence levels:

When constructing confidence intervals, the width is determined by the critical value (z-score) associated with the desired confidence level. Higher confidence levels result in wider intervals, and lower confidence levels produce narrower intervals.

Given:

• The original confidence interval was calculated at a 99.8% confidence level, which has a higher z-score (approximately 2.666).
• An 80% confidence interval will have a lower confidence level and therefore a smaller critical z-score.

For an 80% confidence level, the critical z-score is approximately \(1.28\).

Conclusion:

Since the critical value (z-score) for the 80% confidence interval is smaller than the critical value for the 99.8% confidence interval, the width of the 80% confidence interval will be narrower than the width of the 99.8% confidence interval.

Thus, the answer to your question is that the 80% confidence interval would be narrower than the interval constructed in Part (a).