Sanderson M. Smith

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STATISTICS

HYPOTHESIS TESTING EXAMPLES USING NORMAL DISTRIBUTION================================================

1.

Candidate Jones is one of two candidates running for mayor of Central City. A random polling of 672 registered voters finds that 323 (48% of those polled) will vote for him. It is reasonable to assume that the race for mayor is a tossup? (That is, if p is the proportion of the population who will vote for Jones, is it reasonable to assume that p = 0.5?)

Analysis:The population involved is the registered voters in Central City. As worded, this is a 2-tail situation. We are testing the null hypothesisH

_{o}: p = 0.5against the alternate hypothesis

H

_{a}: p is not equal to 0.5.This is a binomial setting with N = 672 and p = 0.5.

Let p(hat)is the proportion of registered voters in a sample of size 672 who will vote for Jones, then if H

_{o}is true, the distribution of p(hat) is approximately normal, the mean of the distribution of p(hat) is 0.5 and the standard deviation of p(hat) is SQRT[(0.5)(0.5)/672] = 0.019287, or about 1.93%.The mean of our specific sample is 0.48. If H

_{o}is true, the associated z score is (0.475 - 0.5)/0.0193 = -1.3. At the 2-tail, 5% level of significance, the critical z scores are z > 1.96 or z < -1.96. There is not enough evidence to reject H_{o}at the 5% level.[We can also note that normalcdf(-1E99,0.475,0.5,.0193) = 0.0976, or about 10%. Basically, if Ho is true, the probability of obtaining the sample proportion 48% is about 10%. We don't have "strong" evidence to reject H

_{o. }And, we are simply saying that it is not unreasonable to assume that the population parameter is 50%.]============================================

2.

In a quality control situation, the mean weight of objects produced is supposed to be 16 ounces with a standard deviation of 0.4 ounces. A random sample of 70 objects yields a mean weight of 15.8 ounces. Is it reasonable to assume that the production standards are being maintained?

Analysis:The population involved is the means of random samples of size 70 chosen from a population with mean = 16 and standard deviation = 0.4. Call this population P. The Central Limit Theorem says that the distribution of P is normal, that the mean of P is 16, and that the standard deviation of P is 0.4/Ã(70) = 0.048. As worded, this is a 2-tail situation. The null and alternate hypotheses are, respectivelyH

_{o}: The sample came from a population with mean = 16.H

_{a}. The sample did not come from a population with mean = 16.If we test at the 5% level of significance, the critical z scores are z > 1.96 and z < -1.96. The sample z score is z = (15.8-16)/0.048 = -4.17. We therefore reject H

_{o}at the 5% level of significance.[Note: The critical z scores at the 1% level of significance are z < -2.58 and z > 2.58. We would reject Ho at the 1% level of significance. In other words, there is very strong evidence to reject H

_{o}.]========================================================

3.

I roll a single die 1,000 times and obtain a "6" on 204 rolls. Is there significant evidence to suggest that the die is not fair?

Analysis:The population involved consists of the proportion of 6's obtained when a die is rolled 1,000 times. This population consists of 1,001 proportions: {0/1000, 1/1000, 2/1000,É.,999/1000, 1000/1000}. If p is the proportion of 6's obtained, then our null and alternate hypotheses are, respectivelyH

_{o}: p = 1/6H

_{a}: p is not equal to 1/6This is a 2-tail situation. If H

_{o }is true, we have a binomial setting with p = 1/6 and N = 1,000. If p(hat) is a sample proportion of 6's, then the distribution of p(hat is approximately normal, the mean of the distribution of p(hat) is 1/6 and the standard deviation is SQRT[(1/6)(5/6)/1000] = 0.011785, or approximately 0.012.Our sample proportion is p(hat) = 204/1000 = 0.204, or 20.4%. If H

_{o}is true, the probability of obtaining this result is 0.00107, or approximately (1/10)%. [=normalcdf(.2035,1E99,1/6,.012)] There is strong evidence to reject H_{o}, since it is highly unlikely that this result would be obtained if H_{o}is true.[Note: Our z statistics would be z = (0.204-1/6)/.012 = 3.11, which is "way out there" on a normal distribution curve.]

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