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The SAT is the most widely used college admission exam. (Most community colleges do not require students to take this exam.) The mean SAT math score varies by state and by year, so the value of µ depends on the state and the year. But let’s assume that the shape and spread of the distribution of individual SAT math scores in each state is the same each year. More specifically, assume that individual SAT math scores consistently have a normal distribution with a standard deviation of 100. An educational researcher wants to estimate the mean SAT math score (μ) for his state this year. The researcher chooses a random sample of 656 exams in his state. The sample mean for the test is 480.

Find the 95% confidence interval to estimate the mean SAT math score in this state for this year.

(Note: The critical
z
-value to use,
z
c
, is: 1.960.)

( , )

Respuesta :

Using the z-distribution, it is found that the 95% confidence interval to estimate the mean SAT math score in this state for this year is (472, 488).

We have the standard deviation for the population, which is why the z-distribution is used to solve this question.

  • The sample mean is [tex]\overline{x} = 480[/tex].
  • The population standard deviation is [tex]\sigma = 100[/tex].
  • The sample size is [tex]n = 656[/tex].

The interval is given by:

[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]

We have to find the critical value, which is z with a p-value of [tex]\frac{1 + \alpha}{2}[/tex], in which [tex]\alpha[/tex] is the confidence level.

In this problem, [tex]\alpha = 0.95[/tex], thus, z with a p-value of [tex]\frac{1 + 0.95}{2} = 0.975[/tex], which means that it is z = 1.96.

Then:

[tex]\overline{x} - z\frac{\sigma}{\sqrt{n}} = 480 - 1.96\frac{100}{\sqrt{656}} = 472[/tex]

[tex]\overline{x} + z\frac{\sigma}{\sqrt{n}} = 480 + 1.96\frac{100}{\sqrt{656}} = 488[/tex]

The 95% confidence interval to estimate the mean SAT math score in this state for this year is (472, 488).

A similar problem is given at https://brainly.com/question/22596713

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