A researcher would like to determine whether an over-the-counter cold medication

Question

A researcher would like to determine whether an over-the-counter cold medication has an effect on mental alertness. A sample of n = 16 participants is obtained, and each person is given a standard dose of the medication one hour before being tested in a driving simulation task. For the general population, reaction time scores on the simulation task are normally distributed with µ = 210 and s = 20. The individuals in the sample had an average score of M = 222.

A.) Can the research conclude that the medication has significant effect on mental alertness as measured by the diriving simulation task?

B.) Use a two-tailed test with a=.05. Compute Cohens d to measure the size of the effect.

C.)Construct a 90% confidence interval to approximate the mean of the population.

I missed this question on my last stats test and have a final coming up, was wondering if someone could show me how the work it correctly.

Explanation / Answer

PART A.
Given that,
population mean(u)=210
standard deviation, =20
sample mean, x =222
number (n)=16
null, Ho: =210
alternate, H1: !=210
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 222-210/(20/sqrt(16)
zo = 2.4
| zo | = 2.4
critical value
the value of |z | at los 5% is 1.96
we got |zo| =2.4 & | z | = 1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 2.4 ) = 0.0164
hence value of p0.05 > 0.0164, here we reject Ho
ANSWERS
---------------
null, Ho: =210
alternate, H1: !=210
test statistic: 2.4
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.0164

conclude that medication has significant effect on mental alertness as measured by the diriving simulation task

PART C.
TRADITIONAL METHOD
given that,
standard deviation, =20
sample mean, x =222
number (n)=16
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
stanadard error = ( 20/ sqrt ( 16) )
= 5
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.64
since our test is two-tailed
value of z table is 1.64
margin of error = 1.64 * 5
= 8.2
III.
CI = x ± margin of error
confidence interval = [ 222 ± 8.2 ]
= [ 213.8,230.2 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =20
sample mean, x =222
number (n)=16
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.64
since our test is two-tailed
value of z table is 1.64
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 = [ 222 ± Z a/2 ( 20/ Sqrt ( 16) ) ]
= [ 222 - 1.64 * (5) , 222 + 1.64 * (5) ]
= [ 213.8,230.2 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [213.8 , 230.2 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 222
standard error =5
z table value = 1.64
margin of error = 8.2
confidence interval = [ 213.8230.2 ]

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