When a fair die is rolled many times, the outcomes of 1, 2, 3, 4, 5, and 6 are e

Question

When a fair die is rolled many times, the outcomes of 1, 2, 3, 4, 5, and 6 are equally likely, so the mean of the outcomes should be 3.5. After a die is drilled into and loaded with weights, it is rolled 16 times and a mean of 2.9375 is obtained. Assume that the standard deviation of the outcomes is 1.7078, which is the standard deviation for a fair die (the population). Use a 0.05 significance level to test the claim that outcomes from the loaded die have a mean different from the value of 3.5 expected with a fair die.

a.Is there anything about the sample data suggestion that the methods of hypothesis testing used should not be used?

Explanation / Answer

Given that,
population mean(u)=3.5
standard deviation, =1.7078
sample mean, x =2.9375
number (n)=16
null, Ho: =3.5
alternate, H1: !=3.5
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 = 2.9375-3.5/(1.7078/sqrt(16)
zo = -1.31748
| zo | = 1.31748
critical value
the value of |z | at los 5% is 1.96
we got |zo| =1.31748 & | z | = 1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -1.31748 ) = 0.18768
hence value of p0.05 < 0.18768, here we do not reject Ho
ANSWERS
---------------
null, Ho: =3.5
alternate, H1: !=3.5
test statistic: -1.31748
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.18768

claim that outcomes from the loaded die have a mean is no different from the value of 3.5
expected with a fair die

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