According to a survey conducted by the Association for Dressing and Sauces, 85%

Question

According to a survey conducted by the Association for Dressing and Sauces, 85% of Americans adult eat salad once a week. A nutritionist suspects that the percentage is higher at than this. She conducts a survey of 200 American adults and finds that 171 of them eat salad least once a week. Conducts the appropriate test that addresses the nutritionist's suspicions. Use a 0.1 level of significance. Use p-value method A Fair Isaac Corporation (FICO) score is used by credit agencies to assess the creditworthiness of individuals. Its value ranges from 300 to 850. An individual with Fico score over 700 is considered to be a quality credit risk. According to Fairs Isaac Corporation, the mean Fico score is 703.5. A credit analyst wondered whether high-income individuals (incomes in excess of $100,000 per year) had higher credit scores. He obtained a random sample of 40 high- income individuals and found the mean credit score to with standard deviation of 83.2. Conduct the appropriate test to determine if high-income individuals have higher Fico score at the 0.05 significance level. Use p-value method The American Hospital Association reports in Hospital Start that the mean cost to community hospital per patient per day in US. hospital was $931 in 1994. In those same years, a random sample of 30 daily costs in Massachusetts hospital yielded a mean of $1, 131 with the standard deviation of $333 for Massachusetts hospital, do the data provide sufficient evidence to conclude that in 1994 the mean cost in Massachusetts hospital exceeded the national mean of $931? Perform the required hypothesis test at the 5% significant level. Use confidence interval method

Explanation / Answer

Here p0 = 0.85

Null HYpothesis : H0 : Percentage americans eat salad per week p = p0 = 0.85

Alternative Hypotheis : Ha : Precentage Americans ear salad per week. p >  p0

Test statistic

p = 171/ 200 = 0.855

Standard error of sampling se0 = sqrt [p0 (1-p0)/N] = sqrt [0.85 * 0.15/200] = 0.02525

Z = (p - p0)/ se0 = ( 0.855 - 0.85)/ 0.02525 = 0.20

P - value = Pr (p >0.855; 0.85; 0.02525) = 1 - (0.20) = 1- 0.5793 = 0.4207

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