Question

A. A researcher was interested in comparing the GPAs of students at two different colleges. Independent simple random samples of 8 students from college A and 13 students from college B yielded the following results. The mean GPA for college A was x1 = 3.11, with a standard deviation s1 = 0.44. The GPA for college B was 2 = 3.44, with a standard deviation s2 = 0.55. Determine a 95% confidence interval for the difference, µ1-µ2 between the mean GPA of college A students and the mean GPA of college B students. (Assume college a and B have the same population standard deviations)

B. A manufacturing process produces bags of cookies. The distribution of content weights of these bags is Normal with mean 15.0 oz and standard deviation 6.0 oz. We will randomly select a sample of 900 bags of cookies and weigh the contents of each bag selected, is the mean of such sample

a) The mean of is x

b) The standard deviation of x is

c) The distribution of x is

d) Does the distribution of depend on the assumption that the weight (in oz) of these bags is normally distributed and why?

C.Last year, the mean annual salary for adults in one town was $35,000. A researcher wants to perform a hypothesis test to determine whether the mean annual salary for adults in this town has changed this year. The mean annual salary for a random sample of 16 adults from the town was 33000. Assume population standard deviation σ =12000. Use a significance level of α=0.05

D. A special diet is intended to reduce the cholesterol of patients at risk of heart disease. After six months on the diet, an SRS of 64 patients at risk for heart disease had an average cholesterol of x= 192, with standard deviation s = 24. The 95% confidence interval for the average cholesterol of patients at risk for heart disease who have been on the diet for 6 months is

Answer 1

A.

Point estimate of µ1-µ2 = x1 - x2 = 3.11 - 3.44 = -0.33

Pooled variance = Sp^2 = [(n1 - 1) s1^2 + (n2 - 1) s2^2] / (n1 + n2 - 2)

= [(8 - 1) * 0.44^2 + (13 - 1) * 0.55^2] / (8 + 13 - 2)

= 0.2623789

Pooled standard deviation = = 0.51223

Degree of freedom = n1 + n2 - 2 = 8 + 13 - 2 = 19

Critical value of t at 95% confidence interval and df = 19 is, 2.09

95% confidence interval for the difference, µ1-µ2 between the mean GPA of college A students and the mean GPA of college B students is,

(-0.33 - 2.09 * 0.51223 ,  -0.33 + 2.09 * 0.51223)

(-1.400561 ,  0.7405607)

B.

a) The mean of is x = µ = 15.0 oz

b) The standard deviation of x = 6.0 / = 0.2 ox

c) The distribution of x is Normal with mean = 15.0 oz and standard deviation = 0.2 oz

d) The distribution does not depend on the assumption that the weight (in oz) of these bags is normally distributed. According to Central limit theorem, for large sample size, the sampling distribution of mean is normal irrespective of the distribution of the population.

C.

H0:

Ha:

Standard error of mean = 12000 / = 3000

Test statistic, z = (33000 - 35000)/3000 = -0.67

P-value = 2 * P(z < -0.67) = 0.5029

Since, p-value is greater than 0.05 significance level, we fail to reject null hypothesis H0 and conclude that there is no strong evidence that the true mean annual salary for adults in this town has changed this year.

D.

Degree of freedom = 64 - 1 = 63

Critical value of t at 95% confidence interval and df = 63 is 2.00

Standard error of mean = 24 / = 3

95% confidence interval for the average cholesterol of patients at risk for heart disease is,

(192 - 2 * 3,  192 + 2 * 3)

(186 , 198)