Question

1.

Seams Personal advertises on its website that 95% of customer orders are received within four working days. They performed an audit from a random sample of 500 of the 6,000 orders received that month and it shows 470 orders were received on time.

(Question) If Seams Personal customers really receive 95% of their orders within four working days, what is the probability that the proportion in the random sample of 500 orders is the same as the proportion found in the audit sample or less?

2.

You collect a random sample of size n from a population and calculate a 98% confidence interval. Which of the following strategies produces a new confidence interval with a decreased margin of error?

Use a 99% confidence level.  Use a 95% confidence level.  Decrease the sample size.  Use the same confidence level, but compute the interval n times. Approximately 2% of these intervals will be larger.  Nothing can guarantee that you will obtain a larger margin of error. You can only say that the chance of obtaining a larger interval is 0.02.

3.

Faculty members at Lowell Place High School want to determine whether there are enough students to have a Valentine's Day Formal. Eighty-eight of the 200 students said they would attend the Valentine's Day Formal. Construct and interpret a 90% confidence interval for p.

The 90% confidence interval is (0.4977, 0.5023). We are 90% confident that the true proportion of students attending the Valentine's Day Formal is between 49.77% and 50.23%.  The 90% confidence interval is (0.3823, 0.4977). There is a 90% chance that a randomly selected student who will attend the Valentine's Day Formal lies between 38.23% and 49.77%. The 90% confidence interval is (0.4977, 0.5023). Ninety percent of all samples of this size will yield a confidence interval of (0.4977, 0.5023). The 90% confidence interval is (0.3823, 0.4977). Ninety percent of all samples of this size will yield a confidence interval of (0.3823, 0.4977). The 90% confidence interval is (0.3823, 0.4977). We are 90% confident that the true proportion of students attending the Valentine's Day Formal is between 38.23% and 49.77%.

Answer 1

2)

Margin of error = critical value for a confidence level*√p*(1-p)/√n

So margin of error is directly proportional to the confidence level, and inversely proportional to the sample size.

That is if we wll increase the sample size or decrease the confidence level, margin of error will decrease.

So, if we will use 95% confidence level instead of 98, we can reduce the margin of error.

3)

N = 200

P = 88/200

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 88

N*(1-p) = 112

Both the conditions are met so we can use standard normal z table to estimate the interval

Critical value z for 90% confidence level is 1.645

Margin of error(MOE) = z*√p*(1-p)/√n =

MOE = 0.05773926566

Confidence interval is given by

(P - MOE, P+ MOE)

(0.38226073433, 0.49773926566)

The 90% confidence interval is (0.3823, 0.4977). We are 90% confident that the true proportion of students attending the Valentine's Day Formal is between 38.23% and 49.77%.