Suppose that the population of the scores of all high school seniors that took the SAT-M (SAT math) test this year follows a normal distribu

Question

Suppose that the population of the scores of all high school seniors that took the SAT-M (SAT math) test this year follows a normal distribution with unknown population mean and known standard deviation 100. You read a report that says, “On the basis of a simple random sample of 100 high school seniors that took the SAT-M test this year, a confidence interval for population mean is 512.00 ± 25.75.” The confidence level for this interval is

in progress 0
Eliza 18 hours 2021-10-13T03:05:19+00:00 1 Answer 0

Answers ( )

    0
    2021-10-13T03:07:03+00:00

    Answer:

     25.75 = z_{\alpha/2} \frac{100}{\sqrt{100}}

    And solving for the critical value we got:

     z_{\alpha/2}= \frac{25.75*10}{100} = 2.575

    Now we need to find the confidence level and for this case we can use find this probability:

     P(-2.575< Z<2.575)= P(Z<2.575) -P(Z<-2.575)

    And using the normal standard distribution or excel we got:

     P(-2.575< Z<2.575)= P(Z<2.575) -P(Z<-2.575)= 0.9950-0.0050= 0.99

    So then the confidence interval for this case is 99%

    Step-by-step explanation:

    For this case the random variable X is the scores for the SAT math scores and we know that the distribution for X is normal:

     X\sim N(\mu , \sigma =100)

    They select a random sample of n =100 and they construc a confidence interval for the true population mean of interest and they got:

    512.00 \pm 25.75

    for this problem we need know that the confidence interval for the true mean when the deviation is known is given by:

    \bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

    The margin of error is given by:

     ME =z_{\alpha/2} \frac{\sigma}{\sqrt{n}}

    And the margin of error for this interval is  ME = 25.75 then we can solve for the critical value in order to find the confidence level:

     25.75 = z_{\alpha/2} \frac{100}{\sqrt{100}}

    And solving for the critical value we got:

     z_{\alpha/2}= \frac{25.75*10}{100} = 2.575

    Now we need to find the confidence level and for this case we can use find this probability:

     P(-2.575< Z<2.575)= P(Z<2.575) -P(Z<-2.575)

    And using the normal standard distribution or excel we got:

     P(-2.575< Z<2.575)= P(Z<2.575) -P(Z<-2.575)= 0.9950-0.0050= 0.99

    So then the confidence interval for this case is 99%

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )