Question

**Try to answer what you
can.**

**Use the following information
to answer questions 14 - 20**

The principal of a certain high school claims that the SAT Math scores for his students is 660. A random sample of 68 of these students showed a mean 650. Assume that the population standard deviation of the scores is known to be 100. Test using a level of significance of 1% whether the populations mean SAT score for the high school 660.

14. What is the setup for the null and alternative hypothesis?

15. What is the test statistics? What type of test and why?

16. What is the p value?

17. Using the level
of confidence associated with the level of significance given for
the test **(110%-1% = 99%, this means**

**we should find the 99%
confidence interval)**, what is the lower and upper value
for the confidence interval of the mean number of connections?

18. Determine the value of the test statistic.

19. What is/are your
critical (rejection) region(s)? **(You must state it using
critical values and not the p-value method)**

20. Using a = 0.01,
make an **appropriate decision** and draw a
corresponding **valid conclusion** based on your
findings.

Answer #1

Given that,

population mean(u)=660

standard deviation, sigma =100

sample mean, x =650

number (n)=68

null, Ho: μ=660

alternate, H1: μ>660

level of significance, alpha = 0.01

from standard normal table,right tailed z alpha/2 =2.326

since our test is right-tailed

reject Ho, if zo > 2.326

we use test statistic (z) = x-u/(s.d/sqrt(n))

zo = 650-660/(100/sqrt(68)

zo = -0.825

| zo | = 0.825

critical value

the value of |z alpha| at los 1% is 2.326

we got |zo| =0.825 & | z alpha | = 2.326

make decision

hence value of |zo | < | z alpha | and here we do not reject
Ho

p-value : right tail - ha : ( p > -0.825 ) = 0.795

hence value of p0.01 < 0.795, here we do not reject Ho

ANSWERS

---------------

14.

null, Ho: μ=660

alternate, H1: μ>660

15.

test statistic: -0.825

Z test for single mean because population standard deviation is
known.

16.

p-value: 0.795

17.

TRADITIONAL METHOD

given that,

standard deviation, σ =100

sample mean, x =650

population size (n)=68

I.

standard error = sd/ sqrt(n)

where,

sd = population standard deviation

n = population size

standard error = ( 100/ sqrt ( 68) )

= 12.127

II.

margin of error = Z a/2 * (standard error)

where,

Za/2 = Z-table value

level of significance, α = 0.01

from standard normal table, two tailed z α/2 =2.576

since our test is two-tailed

value of z table is 2.576

margin of error = 2.576 * 12.127

= 31.239

III.

CI = x ± margin of error

confidence interval = [ 650 ± 31.239 ]

= [ 618.761,681.239 ]

-----------------------------------------------------------------------------------------------

DIRECT METHOD

given that,

standard deviation, σ =100

sample mean, x =650

population size (n)=68

level of significance, α = 0.01

from standard normal table, two tailed z α/2 =2.576

since our test is two-tailed

value of z table is 2.576

we use CI = x ± Z a/2 * (sd/ Sqrt(n))

where,

x = mean

sd = standard deviation

a = 1 - (confidence level/100)

Za/2 = Z-table value

CI = confidence interval

confidence interval = [ 650 ± Z a/2 ( 100/ Sqrt ( 68) ) ]

= [ 650 - 2.576 * (12.127) , 650 + 2.576 * (12.127) ]

= [ 618.761,681.239 ]

-----------------------------------------------------------------------------------------------

interpretations:

1. we are 99% sure that the interval [618.761 , 681.239 ] contains
the true population mean

2. if a large number of samples are collected, and a confidence
interval is created

for each sample, 99% of these intervals will contains the true
population mean

18.

test statistic: -0.825

19.

critical value: 2.326

20.

decision: do not reject Ho

we do not have enough evidence to support the claim that whether
the populations mean SAT score for the high school 660

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