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
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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 ]
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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 ]
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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