Question

E. canis is a tick-borne disease of dogs that is sometimes contracted by humans. Among infected humans, the distribution of white blood cell (WBC) counts has an unknown mean and standard deviation. In the general population, mean WBC is known to be μ=7250/mm . It is believed that persons infected with E. canis may have significantly different WBC counts, potentially making a WBC test an effective screening for E. canis in humans. In a sample of 15 people infected with E. canis, WBC count was ?=4767/mm with a standard deviation of s = 3204/mm . Answer the questions below. You must state your answers to all components below in a full sentence in a way that someone who isn’t in this class would understand. You may refer to the lectures, Test Yourself questions or the instructor for examples. (10 points)

• a) State the null and alternative hypotheses.

• b) Which statistical test will you use? Provide one reason why.

• c) Determine the appropriate critical value for the test you will use. Use an alpha of 0.05.

• d) Compute your test statistic and compare it to your critical value.

• e) Determine the p-value.

• f) What do you conclude?

• g) Would your conclusions be different had you chosen an alpha of 0.01?

h) By changing your alpha from 0.05 to 0.01 did you increase or decrease the probability of a Type
II error (false negative)?

Answer 1

Given that,
population mean(u)=7250
sample mean, x =4767
standard deviation, s =3204
number (n)=15
null, Ho: μ=7250
alternate, H1: μ!=7250
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.145
since our test is two-tailed
reject Ho, if to < -2.145 OR if to > 2.145
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =4767-7250/(3204/sqrt(15))
to =-3.001
| to | =3.001
critical value
the value of |t α| with n-1 = 14 d.f is 2.145
we got |to| =3.001 & | t α | =2.145
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -3.0014 ) = 0.0095
hence value of p0.05 > 0.0095,here we reject Ho
ANSWERS
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a.
null, Ho: μ=7250
alternate, H1: μ!=7250
b.
T test for single mean with unknown population standard deviation
d.
test statistic: -3.001
c.
critical value: -2.145 , 2.145
decision: reject Ho
e.
p-value: 0.0095
f.
we have enough evidence to support the claim that persons infected with E.canis may have significantly different WBC counts.
g.
Given that,
population mean(u)=7250
sample mean, x =4767
standard deviation, s =3204
number (n)=15
null, Ho: μ=7250
alternate, H1: μ!=7250
level of significance, α = 0.01
from standard normal table, two tailed t α/2 =2.977
since our test is two-tailed
reject Ho, if to < -2.977 OR if to > 2.977
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =4767-7250/(3204/sqrt(15))
to =-3.001
| to | =3.001
critical value
the value of |t α| with n-1 = 14 d.f is 2.977
we got |to| =3.001 & | t α | =2.977
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -3.0014 ) = 0.0095
hence value of p0.01 > 0.0095,here we reject Ho
ANSWERS
---------------
null, Ho: μ=7250
alternate, H1: μ!=7250
test statistic: -3.001
critical value: -2.977 , 2.977
decision: reject Ho
p-value: 0.0095
we have enough evidence to support the claim that persons infected with E.canis may have significantly different WBC counts.
h.
i.
if level of significance =0.05
Given that,
Standard deviation, σ =3204
Sample Mean, X =4767
Null, H0: μ=7250
Alternate, H1: μ!=7250
Level of significance, α = 0.05
From Standard normal table, Z α/2 =1.96
Since our test is two-tailed
Reject Ho, if Zo < -1.96 OR if Zo > 1.96
Reject Ho if (x-7250)/3204/√(n) < -1.96 OR if (x-7250)/3204/√(n) > 1.96
Reject Ho if x < 7250-6279.84/√(n) OR if x > 7250-6279.84/√(n)
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Suppose the size of the sample is n = 15 then the critical region
becomes,
Reject Ho if x < 7250-6279.84/√(15) OR if x > 7250+6279.84/√(15)
Reject Ho if x < 5628.5523 OR if x > 8871.4477
Implies, don't reject Ho if 5628.5523≤ x ≤ 8871.4477
Suppose the true mean is 4767
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(5628.5523 ≤ x ≤ 8871.4477 | μ1 = 4767)
= P(5628.5523-4767/3204/√(15) ≤ x - μ / σ/√n ≤ 8871.4477-4767/3204/√(15)
= P(1.0414 ≤ Z ≤4.9614 )
= P( Z ≤4.9614) - P( Z ≤1.0414)
= 1 - 0.8512 [ Using Z Table ]
= 0.1488
For n =15 the probability of Type II error is 0.1488
ii.
if level of significance =0.01
Given that,
Standard deviation, σ =3204
Sample Mean, X =4767
Null, H0: μ=7250
Alternate, H1: μ!=7250
Level of significance, α = 0.01
From Standard normal table, Z α/2 =2.5758
Since our test is two-tailed
Reject Ho, if Zo < -2.5758 OR if Zo > 2.5758
Reject Ho if (x-7250)/3204/√(n) < -2.5758 OR if (x-7250)/3204/√(n) > 2.5758
Reject Ho if x < 7250-8252.8632/√(n) OR if x > 7250-8252.8632/√(n)
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Suppose the size of the sample is n = 15 then the critical region
becomes,
Reject Ho if x < 7250-8252.8632/√(15) OR if x > 7250+8252.8632/√(15)
Reject Ho if x < 5119.1199 OR if x > 9380.8801
Implies, don't reject Ho if 5119.1199≤ x ≤ 9380.8801
Suppose the true mean is 4767
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(5119.1199 ≤ x ≤ 9380.8801 | μ1 = 4767)
= P(5119.1199-4767/3204/√(15) ≤ x - μ / σ/√n ≤ 9380.8801-4767/3204/√(15)
= P(0.4256 ≤ Z ≤5.5772 )
= P( Z ≤5.5772) - P( Z ≤0.4256)
= 1 - 0.6648 [ Using Z Table ]
= 0.3352
For n =15 the probability of Type II error is 0.3352
if level of significance from 0.05 to 0.01 it will increase the type 2 error.