Please help answering the following questions .. thank you Please help answering

Question

Please help answering the following questions .. thank you Please help answering the following questions .. thank you Please help answering the following questions .. thank you

8. Counselors for high chool" in the Routheast region of Minnotadaired that 25% of their high-school students have tried some form of drugs before the end of their sophomore year. Suppose many sample of 40 students will be drawn from this population and the proportion of these of drugs before the end of their sophomore year is records (a) Use the Central Limit Theorem to describe the sampling distribution of proportion of these students students who have tried some form who have tried some form of drugs before the end of their sophomore year for random samples of 40 students each. (8 noints) (b) Approximately 95% of all sample proportions will fall in between what two values? (7 points) (c) Would a sample proportion of 0.20 from a sample drawn from this population, of stze 40 be unusual? (7 points) Be clear with the answer. 9. In a customer evaluation survey for a national chain of appliance stores, 20 customers were randomly selected to rate their experiences with this chain of stores. The sample mean evaluation given by these customers for the stores on a scale of 1 (low) to 10 (high) was 7.1 with a standard deviation of 2.9 (5 points) (a) Define the population of interest and variable of interest. (5 points) (b) State the point estimate for the mean customer evaluation rating for these stores. (c) Based on the sarnple information, compute and interpret the 95% confidence interval for the mean customer evaluation rating for the stores. Assume that the population has a normal probability distri- (6 points) bution. (d) Suppose a 95% confidence interval with a margin of error of 0.5 is desired. How large of a sample (6 points) should be taken?

Explanation / Answer

PART A.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
mean ( np ) = 40 * 0.26 = 10.4
standard deviation ( npq )= 40*0.26*0.74 = 2.7742
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)

PART B.
LOWER/BELOW
P ( Z < x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is -1.96
P( x-u/s.d < x - 10.4/2.7742 ) = 0.025
That is, ( x - 10.4/2.7742 ) = -1.96
--> x = -1.96 * 2.7742 + 10.4 = 4.9627
UPPER/TOP
P ( Z > x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is 1.96
P( x-u / (s.d) > x - 10.4/2.7742) = 0.025
That is, ( x - 10.4/2.7742) = 1.96
--> x = 1.96 * 2.7742+10.4 = 15.8373
approximately 95% of all sample values falls in b/w 4.9627 and 15.8373
PART C.
P(X < 0.2) = (0.2-0.26)/0.0694
= -0.06/0.0694= -0.8646
= P ( Z <-0.8646) From Standard Normal Table
= 0.1936
since the proportion is not less than 0.05
we consider to be usual

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.