Attempts: 2 Keep the Highest: 2111 s. The sampling distribution of the sample pr
ID: 3222359 • Letter: A
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Attempts: 2 Keep the Highest: 2111 s. The sampling distribution of the sample proportion In 2007, about 30% of new car purchases in California were finanoed with a home equity loan. [Source: Auto Feels of right oedit, The New York May Industry the Pain Times, 27, 2008.1 The ongoing process of new-car purchases in California can be viewed as an infinite population. Define p as the proportion of the population of new car purchases in California that are financed with a home equity loan. Thetrue population value of p is not known, but for the sake of this exerdse assume that p -.30. You will calculate an estimate of p by drawing a random sample of so new car purchases made in California. If the purchase was financed with a home equity loan, you a value of Yes for the variable HomeEquity. the purchase was not financed with a home equity loan, You record a value of No for the variable Home Equity. Since the population is infinite, an infinite number of samples canbe drawn from the population. The sample data for the variable Home Equity for 15 random samples (of size n 50) that could be pulled from the population are inthe data set samples. Each value of Home Equity is a a random selection from a binomial population with p .30.) Data set samples Simple random samples In 50l drawn from a binomial population (p s0.300 G Minitab was used to generate the samples. Observations Variable Form v values Missing Sample 1 Qualitative Sample 2 Qualitative Nonnumeric Non numeric Qualitative Sample 10 Sample 1 Sample 12 Sample 13 Qualitative Non numeric Sample 14 Qualitative Non numeric Non numericExplanation / Answer
the worst case = p = 0.12 or p = 0.48
in both the cases, the proportion is far away from the actual value 0.30
.
the mean value P = 0.30
The standard deviation = 0.06 ......[these values are observed from the histogram provided in the third image]
.
There are 10 samples with one standard deviation , and 14 samples within two standard deivations and 15 samples within 3 standard deviations
I get to know these values , by applyng the empirical rule
68% values lie between 1 standard deivtaion , thus 68% of 15 = 10
95% values lie bteween 2 standard deviations, thus 95% of 15 = 14
99.7% of the values lie between 3 standard deviations , thus 99.7% of 15 = 15
.
Based on the histogram , we would consider the distribution to be normal
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