Question

Laith GhannoomConfidence intervals for MeansMay 09, 10:06:04 PMA study by the department of education of a certain state was trying todetermine the mean SAT scores of the graduating high school seniors.The study examined the scores of a random sample of 51 graduatingseniors and found the mean score to be 512 with a standard deviationof 109. Determine a 95% confidence interval for the mean, rounding allvalues to the nearest tenth.Submit Answer

Answer 1

Confidence interval of the mean.

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=512.

The sample size is N=51.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

`s_M=(s)/(√(N))=(109)/(√(51))=(109)/(7.141)=15.263`

The degrees of freedom for this sample size are:

`df=n-1=51-1=50`

The t-value for a 95% confidence interval and 50 degrees of freedom is t=2.

The margin of error (MOE) can be calculated as:

`MOE=t\cdot s_M=2\cdot15.263=30.526\approx30.5`

Then, the lower and upper bounds of the confidence interval are:

`\begin{gathered} LL=M-t\cdot s_M=512-30.5=481.5 \\ UL=M+t\cdot s_M=512+30.5=542.5 \end{gathered}`

Answer: The 95% confidence interval for the mean is (481.5, 542.5).