Question
Phonics is an instructional method in which children are taught to connect sounds with letters or groups of letters. A sample of 144 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 34.04 with a standard deviation of 23.74.
(a) Construct a 98% confidence interval for the mean number of letter sounds identified in one minute. Round the answers to at least two decimal places.
A 98% confidence interval for the mean number of letter sounds identified in one minute is [blank] < u < [blank]
.
(a) Construct a 98% confidence interval for the mean number of letter sounds identified in one minute. Round the answers to at least two decimal places.
A 98% confidence interval for the mean number of letter sounds identified in one minute is [blank] < u < [blank]
.
Answers
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Answered by
GPT-4o mini
To construct a 98% confidence interval for the mean number of letter sounds identified, we can use the formula for the confidence interval for the mean when the population standard deviation is unknown. We use the t-distribution since we have a sample rather than the entire population.
The formula for the confidence interval is given by:
\[
\bar{x} \pm t_{\alpha/2} \left( \frac{s}{\sqrt{n}} \right)
\]
Where:
- \(\bar{x}\) is the sample mean
- \(t_{\alpha/2}\) is the t-score for the desired level of confidence and degrees of freedom
- \(s\) is the sample standard deviation
- \(n\) is the sample size
### Step 1: Identify the sample statistics
- Sample mean (\(\bar{x}\)) = 34.04
- Sample standard deviation (\(s\)) = 23.74
- Sample size (\(n\)) = 144
### Step 2: Find the t-score
Since this is a 98% confidence interval, we need to find \(\alpha = 1 - 0.98 = 0.02\). The value of \(\alpha/2 = 0.01\).
The degrees of freedom \(df = n - 1 = 144 - 1 = 143\).
Using a t-table or calculator for \(df = 143\) and \(\alpha/2 = 0.01\), we find:
- \(t_{\alpha/2} \approx 2.63\) (You might find a different value in some tables, but 2.63 is a typical approximation for this level of confidence with this degrees of freedom.)
### Step 3: Calculate the standard error
The standard error (SE) can be calculated as:
\[
SE = \frac{s}{\sqrt{n}} = \frac{23.74}{\sqrt{144}} = \frac{23.74}{12} \approx 1.9783
\]
### Step 4: Calculate the margin of error
Now, we calculate the margin of error (ME):
\[
ME = t_{\alpha/2} \cdot SE = 2.63 \cdot 1.9783 \approx 5.20
\]
### Step 5: Construct the confidence interval
Now we can construct the confidence interval:
\[
\text{Lower bound} = \bar{x} - ME = 34.04 - 5.20 = 28.84
\]
\[
\text{Upper bound} = \bar{x} + ME = 34.04 + 5.20 = 39.24
\]
### Final Answer
A 98% confidence interval for the mean number of letter sounds identified in one minute is:
\[
[28.84 < \mu < 39.24]
\]
The formula for the confidence interval is given by:
\[
\bar{x} \pm t_{\alpha/2} \left( \frac{s}{\sqrt{n}} \right)
\]
Where:
- \(\bar{x}\) is the sample mean
- \(t_{\alpha/2}\) is the t-score for the desired level of confidence and degrees of freedom
- \(s\) is the sample standard deviation
- \(n\) is the sample size
### Step 1: Identify the sample statistics
- Sample mean (\(\bar{x}\)) = 34.04
- Sample standard deviation (\(s\)) = 23.74
- Sample size (\(n\)) = 144
### Step 2: Find the t-score
Since this is a 98% confidence interval, we need to find \(\alpha = 1 - 0.98 = 0.02\). The value of \(\alpha/2 = 0.01\).
The degrees of freedom \(df = n - 1 = 144 - 1 = 143\).
Using a t-table or calculator for \(df = 143\) and \(\alpha/2 = 0.01\), we find:
- \(t_{\alpha/2} \approx 2.63\) (You might find a different value in some tables, but 2.63 is a typical approximation for this level of confidence with this degrees of freedom.)
### Step 3: Calculate the standard error
The standard error (SE) can be calculated as:
\[
SE = \frac{s}{\sqrt{n}} = \frac{23.74}{\sqrt{144}} = \frac{23.74}{12} \approx 1.9783
\]
### Step 4: Calculate the margin of error
Now, we calculate the margin of error (ME):
\[
ME = t_{\alpha/2} \cdot SE = 2.63 \cdot 1.9783 \approx 5.20
\]
### Step 5: Construct the confidence interval
Now we can construct the confidence interval:
\[
\text{Lower bound} = \bar{x} - ME = 34.04 - 5.20 = 28.84
\]
\[
\text{Upper bound} = \bar{x} + ME = 34.04 + 5.20 = 39.24
\]
### Final Answer
A 98% confidence interval for the mean number of letter sounds identified in one minute is:
\[
[28.84 < \mu < 39.24]
\]
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