Question

The diameters of a random sample of 20 products made by a certain machine during one month showed a mean of 0.93 inch and a sample standard deviation of 0.05 inch. Assume the population is normally distributed, find a 95% confidence intervals for the mean diameter of all the products.

Select one:

a. 0.9066 to 0.9534 inch

b. 0.9081 to 0.9519 inch

c. 0.9248 to 0.9352 inch

d. 0.9107 to 0.9493 inch

“Is live lecture over Zoom an effective way to conduct online lectures?” A pilot sample of 50 students of UM reveals that 18 of them gave a positive response. If a researcher wants to estimate the proportion of positive responses from all the students of UM, what should be the minimum sample size used if she wants the width of a 95% confidence interval estimate to be 0.10?

Select one:

a. 355

b. 97

c. 385

d. 89

Because of the relatively high interest rates, most consumers attempt to pay off their credit card bills promptly. However, this is not always possible. An analysis of the amount of interest paid monthly by a bank’s Visa cardholders reveals that the amount is normally distributed with a mean of $1,735 and a standard deviation of $285. What interest payment is exceeded by only 15% of the bank’s Visa cardholders?

Select one:

a. $1,585.375

b. $1,438.60

c. $2,031.40

d. $1,884.625

Answer 1

a)

95% Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 20- 1 ) = 2.093
0.93 ± 2.093 * 0.05/√(20)
Lower Limit = 0.93 - 2.093 * 0.05/√(20)
Lower Limit = 0.9066
Upper Limit = 0.93 + 2.093 * 0.05/√(20)
Upper Limit = 0.9534
95% Confidence interval is 0.9066 inch to 0.9534 inch

b)

Sample proportion = 18 / 50 = 0.36

Sample size = Z2/2 * p ( 1 - p) / E2

= 1.962 * 0.36 ( 1 - 0.36) / 0.102

= 88.5

n = 89 (Rounded up to nearest integer)

c)

X ~ N ( µ = 1735 , σ = 285 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.15 = 0.85
To find the value of x
Looking for the probability 0.85 in standard normal table to calculate Z score = 1.04
Z = ( X - µ ) / σ
1.04 = ( X - 1735 ) / 285
X = 2031.4