Question

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 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. $1,884.625

b. $1,438.60

c. $2,031.40

d. $1,585.375

Answer #1

a)

95% Confidence Interval

X̅ ± t(α/2, n-1) S/√(n)

t(α/2, n-1) = t(0.05 /2, 30- 1 ) = 2.045

1.75 ± t(0.05/2, 30 -1) * 0.55/√(30)

Lower Limit = 1.75 - t(0.05/2, 30 -1) 0.55/√(30)

Lower Limit = 1.5446

Upper Limit = 1.75 + t(0.05/2, 30 -1) 0.55/√(30)

Upper Limit = 1.9554

95% Confidence interval is **1.54 to 1.96 meals**

b)

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 = **2031.40**

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

In a recent survey conducted by a professor of UM, 200 students
were asked whether or not they have a satisfying experience with
the e-learning approach adopted by the school in the current
semester. Among the 200 students interviewed, 121 said they have a
satisfying experience. What is the 99% confidence interval for the
proportion of all UM students who have a satisfying experience with
the e-learning approach?
Select one:
a. 0.548 to 0.662
b. 0.524 to 0.686
c. 0.537...

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...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 1 minute ago

asked 3 minutes ago

asked 3 minutes ago

asked 12 minutes ago

asked 41 minutes ago

asked 58 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago