Sanderson M. Smith

STATISTICS
HYPOTHESIS TESTING EXAMPLES USING NORMAL DISTRIBUTION

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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 hypothesis

Ho: p = 0.5

against the alternate hypothesis

Ha: 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 Ho 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 Ho 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 Ho 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 Ho. And, we are simply saying that it is not unreasonable to assume that the population parameter is 50%.]

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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, respectively

Ho: The sample came from a population with mean = 16.

Ha. 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 Ho 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 Ho.]

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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, respectively

Ho: p = 1/6

Ha: p is not equal to 1/6

This is a 2-tail situation. If Ho 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 Ho 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 Ho, since it is highly unlikely that this result would be obtained if Ho 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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