1. The Medassist Pharmaceutical Company makes a pill intended for children susce
ID: 3362171 • Letter: 1
Question
1.The Medassist Pharmaceutical Company makes a pill intended for children susceptible to seizures. The pill is supposed to contain 25.0 mg of its active ingredient. A random sample of 20 pills yielded the amounts (in mg) listed here.
The mean of these observations is 23.285 mg and the standard deviation is 4.534 mg. Assume the data follows a normal distribution. To determine whether or not these pills are acceptable, we test the following at a significance level of = 0.05:
Ho: = 25 against
Ha: 25
a.(2 points) What test is appropriate for answering this question? Explain.
b.(2 points) In the context of this problem, why is it important that this is a two-tailed test?
c.(6 points) Describe a Type I error and its consequences in context.
d.(6 points) Describe a Type II error and its consequences in context.
27.5 26.0 22.9 23.4 23.0 23.9 32.6 20.9 22.9 24.3 24.8 16.1 24.3 17.3 18.9 20.7 33.0 15.6 24.3 23.3Explanation / Answer
a.
we are using t -test for single mean because there are given sample of 20 obervaions we can find sample mean and standard deviation.
Given that,
population mean(u)=25
sample mean, x =23.285
standard deviation, s =4.534
number (n)=20
null, Ho: =25
alternate, H1: !=25
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.093
since our test is two-tailed
reject Ho, if to < -2.093 OR if to > 2.093
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =23.285-25/(4.534/sqrt(20))
to =-1.692
| to | =1.692
critical value
the value of |t | with n-1 = 19 d.f is 2.093
we got |to| =1.692 & | t | =2.093
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.6916 ) = 0.1071
hence value of p0.05 < 0.1071,here we do not reject Ho
ANSWERS
---------------
null, Ho: =25
alternate, H1: !=25
test statistic: -1.692
critical value: -2.093 , 2.093
decision: do not reject Ho
p-value: 0.1071
b. two tailed test appropriate because do not reject the null hypothesis
c.
Given that,
Standard deviation, =4.534
Sample Mean, X =23.285
Null, H0: =25
Alternate, H1: !=25
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-25)/4.534/(n) < -1.96 OR if (x-25)/4.534/(n) > 1.96
Reject Ho if x < 25-8.887/(n) OR if x > 25-8.887/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 20 then the critical region
becomes,
Reject Ho if x < 25-8.887/(20) OR if x > 25+8.887/(20)
Reject Ho if x < 23.013 OR if x > 26.987
Suppose the true mean is 23.285
Probability of Type I error,
P(Type I error) = P(Reject Ho | Ho is true )
= P(23.013 < x OR x >26.987 | 1 = 23.285)
= P(23.013-23.285/4.534/(20) < x - / /n OR x - / /n >26.987-23.285/4.534/(20)
= P(-0.268 < Z OR Z >3.651 )
= P( Z <-0.268) + P( Z > 3.651)
= 0.3943 + 0.0001 [ Using Z Table ]
= 0.394
d.
Given that,
Standard deviation, =4.534
Sample Mean, X =23.285
Null, H0: =25
Alternate, H1: !=25
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-25)/4.534/(n) < -1.96 OR if (x-25)/4.534/(n) > 1.96
Reject Ho if x < 25-8.8866/(n) OR if x > 25-8.8866/(n)
-----------------------------------------------------------------------------------------------------
Suppose the size of the sample is n = 20 then the critical region
becomes,
Reject Ho if x < 25-8.8866/(20) OR if x > 25+8.8866/(20)
Reject Ho if x < 23.0129 OR if x > 26.9871
Implies, don't reject Ho if 23.0129 x 26.9871
Suppose the true mean is 23.285
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(23.0129 x 26.9871 | 1 = 23.285)
= P(23.0129-23.285/4.534/(20) x - / /n 26.9871-23.285/4.534/(20)
= P(-0.2684 Z 3.6516 )
= P( Z 3.6516) - P( Z -0.2684)
= 0.9999 - 0.3942 [ Using Z Table ]
= 0.6057
For n =20 the probability of Type II error is 0.6057
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.