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**

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. 0.9107 to 0.9493 inch
b. 0.9066 to 0.9534 inch
c. 0.9081 to 0.9519 inch
d. 0.9248 to 0.9352 inch

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. $2,031.40
b. $1,884.625
c. $1,438.60
d. $1,585.375

A survey of 30 couples showed the sample mean number of meals
eaten outside their home was 1.75 meals per week, with a standard
deviation of 0.55 meal per week. Construct a 95% confidence
interval for the population mean number of meals eaten outside per
week.
Select one:
a. None of the other choices
b. 1.72 to 1.78 meals
c. 1.58 to 1.92 meals
d. 1.54 to 1.96 meals
Because of the relatively high interest rates, most consumers
attempt to...

“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. 385
B. 97
C. 355
D. 89

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adopted the book. Fifteen percent of the professors who did not
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the book, what is the probability that he has received the
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a. 0.2500
b. 0.9375
c. 0.0375
d. 0.2250
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