Question 4 The following Data represents grade point averages of 10 student-samp

Question

Question 4

The following Data represents grade point averages of 10 student-sample taken randomly from a population of students:

Student Grade Average: 2.2 , 2.4 , 2.1 , 1.9 , 2.3 , 2.2 , 2.3 , 2.3 , 2.3 , 2.1

In establishing a confidence interval on the population mean of GPA at 99%, the margin of error (%) is (three decimal places) (when i do this i get 6.742 please explain how to do this)

A claim was made of strict grading in a new math course. The claim indicates that students in this course make a point average of 2.3. Suppose you know that the standard deviation of population GPA is 0.3, test this claim at 5% type 1 error using null hypothesis Ho: µ = 2.3 and answer the following question: The test statistics (three Decimal Places) is:

Explanation / Answer

4.

DIRECT METHOD
given that,
sample mean, x =2.21
standard deviation, s =0.1449
sample size, n =10
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 3.25
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 2.21 ± t a/2 ( 0.1449/ Sqrt ( 10) ]
= [ 2.21-(3.25 * 0.046) , 2.21+(3.25 * 0.046) ]
= [ 2.061 , 2.359 ]
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 9 d.f is 3.25
margin of error = 3.25 * 0.046
= 0.149 = 14.9%

Given that,
Standard deviation, =0.3
Sample Mean, X =2.21
Null, H0: =2.3
Alternate, H1: !=2.3
Level of significance, = 0.05
From Standard normal table, Z /2 =1.96
Since our test is two-tailed
Reject Ho, if Zo < -1.96 OR if Zo > 1.96
Reject Ho if (x-2.3)/0.3/(n) < -1.96 OR if (x-2.3)/0.3/(n) > 1.96
Reject Ho if x < 2.3-0.588/(n) OR if x > 2.3-0.588/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 10 then the critical region
becomes,
Reject Ho if x < 2.3-0.588/(10) OR if x > 2.3+0.588/(10)
Reject Ho if x < 2.114 OR if x > 2.486
Suppose the true mean is 2.3
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(2.114 < x OR x >2.486 | 1 = 2.3)
= P(2.114-2.3/0.3/(10) < x - / /n OR x - / /n >2.486-2.3/0.3/(10)
= P(-1.961 < Z OR Z >1.961 )
= P( Z <-1.961) + P( Z > 1.961)
= 0.0249 + 0.0249 [ Using Z Table ]
= 0.05

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