Q:

# 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) ?

Accepted Solution

A:
Answer:a) $$34.06-2.355\frac{23.83}{\sqrt{134}}=29.212$$    $$34.06+2.355\frac{23.83}{\sqrt{134}}=38.908$$    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. Step-by-step 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 aThe confidence interval for the mean is given by the following formula: $$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$$   (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:$$34.06-2.355\frac{23.83}{\sqrt{134}}=29.212$$    $$34.06+2.355\frac{23.83}{\sqrt{134}}=38.908$$    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.