Thursday, August 27, 2015

Yes the CLT is a bit of a confusion here because as we know, The central limit theorem states that the sampling distribution of any statistic will be normal or nearly normal, if the sample size is large enough. That's where the problem comes in, sample size should be at least 30, some say 40. That would lead you to believe that you cannot use apply CLT here, BUT. the more closely the original population resembles a normal distribution, the fewer sample points will be required.

But what we can do is use the normal approximation to the binomial if this condition holds true. If np > 5 and n(1-p)> 5, then it can be used. We know p = .5 and if you consider 4 trials, then np < 5, as is n(1-p). BUT if you use all 12 tosses, then n = 12 and np = 6 and n(1-p) = 6. But doing it this was we would get the average number of heads expected in 12 tosses to be 6 and the average in all 12 tosses to be 8, instead of 2 in 3 tosses.

npq = (12)(.5)(.5) = 3
sqrt(npq) = 1.732
Formula is z = (x-bar - np)/(sqrt(npq)
z= (8 - 6)/(1.732) = 1.15
z(1.15) = .8749
1- .8749 = .1251
If you notice when you go through a coin tossing experiment tossing three times, there are 8 outcomes and the only way to get greater the 2 heads is if you get HHH, which is 1 out of 8, which is 12.5%.

Thursday, August 20, 2015

since population standard deviation is known, we can use Z

Test stat Z = (x-bar - mean)/(standard deviation/sqrt(n))
Z = (16.2 - 15)/(5.6/sqrt(49))

If you calculate this correctly , you should get Z = 1.50

The critical values are Z = 2.43 and Z = -2.43 since this is a two-tailed test. The critical value was found looking up .9925 on the chart. Took .015/2 = .0075 and then 1-. 0075 to get .9925. It was alpha/2 because of the two-tailed test.

Since 1.5 < 2.43, accept Ho

P-value was .133614. Using the chart it's .1336. In any event, since that value is greater than alpha for the test, accept Ho

Saturday, August 15, 2015

The power of the test is probability of Type II error. IN this case it is probability of type II error when x-bar = 23.6. This is the probability of Ho not being rejected when it is false because x-bar = 23.6
B(23.6) = 1 - z(23.6 - 25.1)(14/sqrt(100))
1- z(-1.07)
Finr z(-1.07) on standard normal distribution chart to get .1423
1- .1423 = .8577
b) we know the power of the test is .9
so 1 - z(23.6 - 25.1)/(14/sqrt(n)) = .9
therefore z(-1.5/914/sqrt(n)) = .1
looking at the z-chart we know z(-1.28) = .1
so -1.28 = -1.5(14/sqrt(n))
-17.92/sqrt(n) = -1.5
cross multiply and square both sides and solve for n and you get n = 143.

Sunday, August 9, 2015

Chart

NY             NJ             CT

Male                     100             60               50
Female                   80             50               80

1) Probability person is female and lives in NY

P(F and NY)

80 females from New York and (100 + 60 + 50 + 80 + 50 + 80) = 420 people total in the company from those three states

P(F and NY) = 80/420 = 4/21

2) Probability person is male or lives in NJ

P(M or NJ) = (100 + 60 + 50 + 50)/420

= 260/420

= 13/21

3)  P(CT if M)

Number of male and living in Connecticut = 50
Number of males (100 + 60 + 50) = 210

P(CT if M) = 50/210 = 5/21

Thursday, August 6, 2015

If a random sample of 28 homes south of Center Street in Provo has a mean selling price of \$145,450 and a standard deviation of \$4775, and a random sample of 25 homes north of Center Street has a mean selling price of \$148,300 and a standard deviation of \$5900, can you conclude that there is a significant difference between the selling price of homes in these two areas of Provo at the 0.05 level?

Ho: Mean1 = Mean2
Ha: Mean 1 does not equal Mean2
x-bar1 (south of Center Street) = 145,500 s1 = 4775, n1 = 28
x-bar2 (north of Center Street) = 148,300 s2 = 5900, n2 = 25
test statistic for the test is (x1-bar - x2-bar)/sqrt(s1^2/n1 + s2^2/n2)
If you put the values into the formula and calculate correctly, you should get
t= -1.919
Make sure you understand how to get that
P-value was obtained from this site. http://www.socscistatistics.com/pvalues/tdistribution.aspx
Just enter the t-score of -1.919, 24 df two sides test and alpha = .05 and you get
.0670
b) the conclusion, since .0670 > .05, we have to accept Ho. So you fail to reject Ho since there is not significant evidence of a difference of means.
Remember when p-value > alpha, accept Ho and when p-value < alpha, reject Ho.