Answer to Question #320137 in Statistics and Probability for Ali De Austria

Question

Computing for the Length of the Confidence Interval

Solve the following.

1. A random sample of size 120 has a standard deviation of 25. What is the margin of error

for a 90% confidence level?

2. Compute the length of the confidence interval with a sample size of 220 and with a standard

deviation 60 having an 80% confidence level.

3. The average price of 50 Philippine-made shirts is PIs150 and the population standard

deviation is Pts55. Find the length of the confidence interval of the true average price of the

shirts using a 95% confidence level.

Answer 1

n =120

s=25

90% confidence level is Z$\alpha$ =1.645

Margin of Error=Z$\alpha$ $\cdot$ s/$\surd$ n

Margin of Error=1.645 25/$\surd$120

Margin of Error=3.7542

2.

n=220

s=60

80% confidence level is Z$\alpha$ =1.282

Length of confidence interval=2$\cdot$ margin of error

Margin of Error=Z$\alpha$ $\cdot$ s/$\surd$ n

Margin of Error=1.282$\cdot$60/$\surd$220

Margin of Error=5.1779

Length of confidence interval =10.3558

3.

n=50

mean =150

$\sigma$=55

95% confidence level is Z$\alpha$ =1.96

Length of confidence interval=2$\cdot$ margin of error

Margin of Error=Z$\alpha$ $\cdot$ s/$\surd$ n

Margin of Error=1.96$\cdot$55/$\surd$50

Margin of Error=15.2452

Length of confidence interval =30.4904