In the town of Springdale, 30% of the families p=.30 regularly plant vegetable gardens. A random sample?

In the town of Springdale, 30% of the families p=.30 regularly plant vegetable gardens. A random sample of n=100 families is selected for a particular study. What is the probability that the sample proportion, the proportion of families who plant a vegetable garden, is between .20 and .40





a. .0146


b. .0292


c. .4854


d. .9708


e. None of the above.





The U.S Bureau of the census reported mean hourly earnings in the wholesale trade industry of u=9.70 per hour in 1987. A random sample of n=49 wholesale trade workers in a particular city showed a sample mean hourly wage of x=9.30 per hour with a standard deviation of s=1.05 per hour. Test to see if wage rates in the city differ significantly from the reported 9.70.

In the town of Springdale, 30% of the families p=.30 regularly plant vegetable gardens. A random sample?
Let X be the number of families that plant vegetable gardens. X has the binomial distribution with n = 100 trials and success probability p = 0.30





In general, if X has the binomial distribution with n trials and a success probability of p then


P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)


for values of x = 0, 1, 2, ..., n


P[X = x] = 0 for any other value of x.





this is found by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.


Or, to be more accurate, the binomial is the sum of n independent and identically distributed Bernoulli trials.





the mean of the binomial distribution is n * p


the variance of the binomial distribution is n * p * (1 - p)





Find P( 20 ≤ X ≤ 40) = 0.9786144








to use the normal approximation to the binomial you need to have at least 10 expected failures and at least 10 expected successes.


expected successes are n * p %26gt; 10


expected failures are n * (1 - p) %26gt; 10





don't forget to use the continuity correction. Draw the histogram of the binomial and shade in the boxes you need to help make sure you do this correctly.





Xn is the normal approximation


Xn ~ Normal(μ = n * p, σ = sqrt(n * p * (1-p)))





Xn ~ Normal(μ = 30, σ = sqrt(21))





Find P( 20 ≤ X ≤ 40)


≈ P(19.5 %26lt; Xn %26lt; 40.5)


= P( (19.5 - 30) / sqrt(21) %26lt; Z %26lt; (40.5 - 30) / sqrt(21))


= P(-2.291288 %26lt; Z %26lt; 2.291288)


= P( Z %26lt; 2.291288) - P( Z %26lt; -2.291288)


= 0.98902662 - 0.01097338


= 0.9780532





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Hypothesis Test for mean:





assuming you have a large enough sample such that the central limit theorem holds and the mean is normally distributed then to test the null hypothesis H0: μ = Δ





find the test statistic z = (xBar - Δ) / (sx / sqrt(n))





where xbar is the sample average


sx is the sample standard deviation


n is the sample size





The p-value of the test is the area under the normal curve that is in agreement with the alternate hypothesis.





H1: p %26gt; p0; p-value is the area to the right of z


H1: p %26lt; p0; p-value is the area to the left of z


H1: p ≠ p0; p-value is the area in the tails greater than |z|





the test statistic for the US Bureau question is:





z = (9.30 - 9.70) / (1.05 / sqrt(49))


z = -2.666667





p-value = P( |Z| %26gt; 2.666667) = 0.007660754





with a low p-value we can reject the null hypothesis and conclude the alternate is true that wage is different.

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