Compute the probability beta of computing a Type II error if the braking distanc

Question

Compute the probability beta of computing a Type II error if the braking distance with the improved tires is now 155 feet.

Determine the number of tests required if it’s necessary that 155 0.01

Ex: The average braking distance from 60 mph of a Mercury Sable is 159 feet. Sables equipped with (hopefully) improved tires have just undergone early testing; the results of the first 45 tests are shown below. Does this indicate that the new tires have decreased the braking distance? Use 0.05. 39.6 170.4 127.9 157.7 172.4 175.4 182.0 157.6 125.8 50.9 157.1 175.4 174.2 136.4 171.1 147.3 137.8 167.6 151.1 159.1 120.1 143.2 143.6 171.8 151.0 146.0 120.3 134.9 150.8 178.8 179.2 126.2 125.5 159.4 132.2 118.2 20.9 165.0 164.4 183.0 175.2 175.5 167.7 163.2 180.0

Explanation / Answer

PART A>

Given that,
population mean(u)=159
sample mean, x =154.0644
standard deviation, s =19.9405
number (n)=45
null, Ho: =159
alternate, H1: <159
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.68
since our test is left-tailed
reject Ho, if to < -1.68
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =154.0644-159/(19.9405/sqrt(45))
to =-1.6604
| to | =1.6604
critical value
the value of |t | with n-1 = 44 d.f is 1.68
we got |to| =1.6604 & | t | =1.68
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :left tail - Ha : ( p < -1.6604 ) = 0.05197
hence value of p0.05 < 0.05197,here we do not reject Ho
ANSWERS
---------------
null, Ho: =159
alternate, H1: <159
test statistic: -1.6604
critical value: -1.68
decision: do not reject Ho
p-value: 0.05197
no evidence to support the new tries are decreased the breaking distance

PART B.

Given that,
Standard deviation, =19.9405
Sample Mean, X =154.0644
Null, H0: =159
Alternate, H1: !=159
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-159)/19.9405/(n) < -1.96 OR if (x-159)/19.9405/(n) > 1.96
Reject Ho if x < 159-39.083/(n) OR if x > 159-39.083/(n)
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Suppose the size of the sample is n = 45 then the critical region
becomes,
Reject Ho if x < 159-39.083/(45) OR if x > 159+39.083/(45)
Reject Ho if x < 153.174 OR if x > 164.826
Implies, don't reject Ho if 153.174 x 164.826
Suppose the true mean is 155
Probability of Type II error,
P(Type II error) = P(Don't Reject Ho | H1 is true )
= P(153.174 x 164.826 | 1 = 155)
= P(153.174-155/19.9405/(45) x - / /n 164.826-155/19.9405/(45)
= P(-0.614 Z 3.306 )
= P( Z 3.306) - P( Z -0.614)
= 0.9995 - 0.2696 [ Using Z Table ]
= 0.73
For n =45 the probability of Type II error is 0.73

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