Question

1. The average of a random sample of 10 scores on a college placement exam is x ¯= 75, and the sample standard deviation is s = 8.4. Assuming that the collection of all scores on the exam is approximately normally distributed, find a 90 percent confidence interval for the mean score.

2. In a random sample of 100 eligible voters, 85 preferred candidate A over candidate B. Find a 99 percent confidence interval for the proportion p of voters favoring candidate A.

Answer 1

Solution :

Given that,

1.

t /2,df = 1.833

Margin of error = E = t/2,df * (s /n)

= 1.833 * (8.4 / 10)

Margin of error = E = 4.9

The 90% confidence interval estimate of the population mean is,

- E < < + E

75 - 4.9 < < 75 + 4.9

70.1 < < 79.9

(70.1 , 79.9)

2.

Point estimate = sample proportion = = x / n = 85 / 100= 0.850

1 - = 1 - 0.850 = 0.15

Z/2 = 2.576

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.576 * (((0.850 * 0.15) / 100)

Margin of error = E = 0.092

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.850 - 0.092 < p < 0.850 + 0.092

0.758 < p < 0.942

The 99% confidence interval for the population proportion p is : 0.758 . 0.942