Question

Phonics is an instructional method in which children are taught to connect sounds with letters or groups of letters. A sample of 134 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 34.06 with a standard deviation of 23.83.

a. Construct a 98% confidence interval for the mean number of letter sounds identified in one minute.
b. If a 95% confidence interval were constructed with these data, would it be wider or narrower than the interval constructed in part (a)? Explain.

Answer 1

a)
`34.06-2.355(23.83)/(√(134))=29.212`

`34.06+2.355(23.83)/(√(134))=38.908`

b) For this case since the confidence level decrease then the confidence interval would decrease respect a) also since the margin of error would be lower.

Explanation:

Information given

`\bar X=34.06` represent the sample mean

`\mu` population mean (variable of interest)

s=23.83 represent the sample standard deviation

n=134 represent the sample size

Part a

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

The degrees of freedom are given:

`df=n-1=134-1=133`

The Confidence level is 0.98 or 98%, the value of significance is
`\alpha=0.02` and
`\alpha/2 =0.01`, and the critical value would be
`t_(\alpha/2)=2.355`

Replacing we got:

`34.06-2.355(23.83)/(√(134))=29.212`

`34.06+2.355(23.83)/(√(134))=38.908`

Part b

For this case since the confidence level decrease then the confidence interval would decrease respect a) also since the margin of error would be lower.