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 27, 2015

## Thursday, August 20, 2015

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

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

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

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.

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