Question

TEST or PARAMETER you would use. NEW! Before SOLVING each problem, state the statistical topic, formula/TEST or PARAN Then SOLVE each Problem! NEW! 1. Scientists conducted an experiment in 2008 to determine side effects of a new drug. Before starting the weight, age, height, allergies, and gender. The scientists gave half of the patients a placebo (fake drug) and the on months the patients reported back and their weights were reevaluated. Moreover, the scientists asked about any side all variables and type of each for this experiment. new drug. Before starting the tests, they recorded the patients' starting ke drug) and the other half the real drug. After two ked about any side effects the patients felt. List 2. Two machines fill 32 oz cartons with cereal. The quality control manager claims that cartons filled by machine I have more filled by machine 2. Samples from each line were taken and the following results were obtained claims that cartons filled by machine I have more cereal than those Machine 1 Machine 2 sample size 25 sample mean 31.8 oz 30.9 oz sample st. dev. 2.2 oz 1.98 oz 36 At the 0.02 significance level, test the claim that the population mean fill for machine is larger than the population mean till for machine 2 3. A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 204 mg with s = 13.3 mg. Construct a 95% confidence interval for the true mean cholesterol content of all such eggs. 4. Use a Chi-Square Test to test the claim that in the given contingency table, the row variable and the column variable are independent. The table below shows the age and favorite type of music of 668 randomly selected people: Rock ор Classic 73 Age 15 to 25 25 to 35 35 to <45 08 77 Use a 5% level of significance to test the null hypothesis that age and preferred music type are independent.

Answer 1

1. The variables are -

i) weight - numerical and continuous

ii) age -numerical and continuous

iii) height - numerical and continuous

iv) allergies - nominal

v) gender - nominal

vi) side effects - nominal

2.

u - u2 = 0 against alternative hypothesis H u -u2 > 0 To test null hypothesis H Here, C1 = 31.800, 1 = 2.200, n1 = 25, 22 = 30.900, S2 = 1.980, n2 = 36 The test statistic can be written as: +_ (C1 – 12) – 0 t= 4 which under Ho follows at distribution with 48 df. where Decision rule : We reject Ho at 0.02 level of significance if P-value < 0.02 or if tobs > to.02,df Now, The value of the test statistic = 1.636364 associated degrees of freedom = 48 The critical value =2.111073 P – value = P(t48 > tobs) = P(t48 > 1.636364) = P(t48 > 1.636364) = 1 – 0.945848 = 0.054152 Since p-value > 0.02 and tobs * tcritical = 2.111073, so We fail to reject Ho at 0.02 level of significance Hence, we can conclude that mean for first population is not significantly greater than the mean for second population.

3.

Let I denotes the sample mean, s denotes the sample standard deviation and n denotes the sample size. Here, Ž = 204.00, s2 = 176.8900, s = 13.30, n = 12 The margin of error is: 13.300 ME= 13.300 - - *t 1-0.950 -0.950 = * t *2.200985 = 8.450417 ~ 8.450 11 V12 * 10.025,11 = V12 a 95 percent confidence interval population mean u is SS * t 1-0.950 2 ,11 13.300 = 204.000 + * 10.025,11 V12 = 204.000 + 13.300 - * 2.200985 V12 = 204.000 – 8.450417 = (195.549583, 212.450417) (195.55, 212.45)