Question

1. A certain town has 9,000 families.Population average mileage driven per family is 15,000 miles per year and the population SD is 2,000 miles per year. Fifteen percent of these families have no cars at all. As part of an opinion survey, a simple random sample of 900 families (from this town) is chosen. What is the chance that sample average mileage driven per family is between 14,950 and 15,100 miles per year? Use the normal approximation method with continuity correction.

2. Somebody picks one ticket at random from a normal population (sample size is equal to 1). The ticket shows a “2”, i.e. the sample average is “2”. Assuming that the SD of the box is 3, determine the 95% confidence interval for the average of the box. (Remark: You should be able to construct a z or t confidence interval.)

3. Determine a 99% confidence interval for the population average of a normal distribution, given a random sample of size 12 with sample average = 2 and sample SD = 3.

Answer 1

1.

n = 900

= 15000 miles

= 2000 miles

Standard error of mean , SE = / = 2000 / = 66.67

Using Normal approximation, that sample average mileage driven per family is between 14,950 and 15,100 miles per year is,

P(14950 < X < 15100) = P(X < 15100) - P(X < 14950)

= P[Z < (15100 - 15000)/66.67] - P[Z < (14950 - 15000)/66.67]

= P[Z < 1.5] - P[Z < -0.75]

= 0.9332 - 0.2266

= 0.7066

2.

As, we know the population standard deviation, we will use Z confidence interval.

Z value for 95% confidence interval is 1.96

95% confidence interval for the average of the box is,

(2 - 1.96 * 3 ,  2 + 1.96 * 3)

(-3.88 , 7.88)

As, the number on ticket cannot be negative, the  95% confidence interval for the average of the box is,

(0, 7.88)

3.

As, we are using the sample standard deviation, we will use t confidence interval.

Degree of freedom = n-1 = 12-1 = 11

Critical value of t at 99% confidence interval and df = 11 is 3.11

Standard error of mean, SE = s / = 3 / = 0.8660254

95% confidence interval for the average of the box is,

(2 - 3.11 * 0.8660254 ,  2 + 3.11 * 0.8660254)

(-0.69 , 4.69)

As, the number on ticket cannot be negative, the  99% confidence interval for the average of the box is,

(0, 4.69)