2. (15pts.) Pennsylvania Lottery. In Pennsylvania Match 6 Lottery, six numbers b

Question

2. (15pts.) Pennsylvania Lottery. In Pennsylvania Match 6 Lottery, six numbers between 1 and 49 are randomly drawn. Use a calculator or a computer to generate 100 random numbers between 1 and 49 (with replacement) and calculate the mean and the standard deviation of your sample. Use a 0.01 significance level to test the claim that the sample is selected from a population with a mean equal to 25, which is the mean of the population of all drawn numbers. Include a copy of the sample you are using.

Explanation / Answer

2.

samples 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,454,46,47,48,49
by calculator or computer
sample mean, x =33.346
standard deviation, s =62.919
Given that,
population mean(u)=25
sample mean, x =33.346
standard deviation, s =62.919
number (n)=100
null, Ho: =25
alternate, H1: !=25
level of significance, = 0.01
from standard normal table, two tailed t /2 =2.6264
since our test is two-tailed
reject Ho, if to < -2.6264 OR if to > 2.6264
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =33.346-25/(62.919/sqrt(100))
to =1.326
| to | =1.326
critical value
the value of |t | with n-1 = 99 d.f is 2.6264
we got |to| =1.326 & | t | =2.6264
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 1.3265 ) = 0.1877
hence value of p0.01 < 0.1877,here we do not reject Ho
ANSWERS
---------------
null, Ho: =25
alternate, H1: !=25
test statistic: 1.326
critical value: -2.6264 , 2.6264
decision: do not reject Ho
p-value: 0.1877
we do not have enough evidence to support the claim

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