Question

the grade point average collected from a random sample of 150 students producing a mean of 2.86. assume that the population standard deviation is 0.78. construct the confidence interval for the population mean, μ, if c = 0.98.

a. (2.71,3.01)
b. (2.51,3.53)
c. (2.31,3.88)
d. (2.43,3.79)

Explanation:

Step1: Find the z - score

For a confidence level \(c = 0.98\), the significance level \(\alpha=1 - c=1 - 0.98 = 0.02\). Then \(\frac{\alpha}{2}=0.01\).
Looking up in the standard normal distribution table, \(z_{\frac{\alpha}{2}}=z_{0.01}\approx 2.33\) (using the property of the standard normal distribution \(P(Z>z_{\frac{\alpha}{2}})=\frac{\alpha}{2}\)).

Step2: Calculate the margin of error \(E\)

The formula for the margin of error when the population standard deviation \(\sigma\) is known is \(E = z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt{n}}\).
Given \(n = 150\), \(\sigma=0.78\), and \(z_{\frac{\alpha}{2}} = 2.33\).

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Step3: Calculate the confidence interval

The formula for the confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu <\bar{x}+E\).
Given \(\bar{x}=2.86\) and \(E = 0.15\).
\(\bar{x}-E=2.86 - 0.15=2.71\) and \(\bar{x}+E=2.86+0.15 = 3.01\)

Answer:

A. \((2.71,3.01)\)