A soft drink manufacturer wishes to know how many soft drinks teenagers drink each week. They want to construct a 98% confidence interval for the mean and are assuming that the population standard deviation for the number of soft drinks consumed each week is 1.3. The study found that for a sample of 2549 teenagers the mean number of soft drinks consumed per week is 5.4.
Construct the desired confidence interval. Round your answers to one decimal place.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

[tex]\bar X=5.4[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma=1.3[/tex] represent the sample standard deviation

n=2549 represent the sample size

Solution to the problem

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

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

Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.01,0,1)".And we see that [tex]z_{\alpha/2}=2.326[/tex]

Now we have everything in order to replace into formula (1):

[tex]5.4-2.326\frac{1.3}{\sqrt{2549}}=5.340[/tex]

[tex]5.4+2.326\frac{1.3}{\sqrt{2549}}=5.460[/tex]

So on this case the 98% confidence interval would be given by (5.3;5.5) after round