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. (c) If a sample of 150 students had been studied, would you expect the confidence interval to be wider or narrower than the interval constructed in part (a) ?

Answer:a) [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]    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 aThe 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 bFor this case since the confidence level decrease then the confidence interval would decrease respect a) also since the margin of error would be lower.