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