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.

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.

Step-by-step explanation:

Information given

[tex]\bar X=34.06[/tex] represent the sample mean

[tex]\mu[/tex] 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:

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

The degrees of freedom are given:

[tex]df=n-1=134-1=133[/tex]

The Confidence level is 0.98 or 98%, the value of significance is [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and the critical value would be [tex]t_{\alpha/2}=2.355[/tex]

Replacing we got:

[tex]34.06-2.355\frac{23.83}{\sqrt{134}}=29.212[/tex]

[tex]34.06+2.355\frac{23.83}{\sqrt{134}}=38.908[/tex]

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.