Statistics Assignment Solution

Solutions: Assignment 2

Please Note:        This assignment is due in class on Wednesday, August 7, 2013. We will    collect hard copies of the assignment

You may do the assignment either individually or a group of not more than two persons. If you do the assignment in a pair please make sure that you only submit one assignment for both the persons, however make sure both names and sections are written on the submitted assignment.

Please submit a single printed word file. If you wish to include excel sheets please copy and paste the sheet in this single word file. Please write your name(s) on the submitted assignment.

This second homework assignment (7 problems) gives you practice constructing and interpreting confidence intervals and conducting various types of Hypothesis Tests.  

Problem 1

A department store is interested in the average balance that is carried on its store’s credit card.  A sample of 40 accounts reveals an average balance of $1,250 and a standard deviation of $350.

Find a 95% confidence interval for the mean account balance on this store’s credit card

Answer

s = 350  

[pic 1]= 1250

α = 0.05  

n = 40

tα/2 = TINV(0.05,39) = 2.0227

The confidence interval is

[pic 2]           i.e.  $1,138.06 < μ  < $1,361.93

Problem 2

You are trying to estimate the average amount a family spends on food during a year.  In the past, the standard deviation of the amount a family has spent on food during a year has been approximately $1200 (assume $1200 to be the population standard deviation). You will be surveying many families and then noting down their amount spent on food. You will then take the average expenditure across all these families to try and approximate the true (population) average family food expenditure. If you want to be 99% sure that you have estimated the true average family food expenditures within plus/minus $60 of your sample mean, how many families do you need to survey?    

Answer

        σ =1200  

        α = 0.01  

        zα/2 = NORMINV(0.005,0,1) = -2.5758

        The confidence interval is

        [pic 3]

        However, we are given that we want the interval to be plus/minus $60 of the sample mean. Thus,

        [pic 4]

        Hence [pic 5]

        Thus we need to survey at least 2654 families

Problem 3

Auditors of Independent Bank are interested in comparing the reported value of all the 1775 customer saving account balances (the bank has 1775 savings accounts) with their own findings regarding the actual value of such assets.  Rather than reviewing the records of each savings account at the bank, the auditors randomly selected a sample of 100 savings account balances.  The sample mean and sample standard deviations were $505.75 and 360.95, respectively.  

(a)        Construct a 90% confidence interval for the total value of all savings account balances within this bank.    

(b)        Interpret the 90% confidence interval constructed in part (a).

Answer

(a)

s = 360.95  

[pic 6]= 505.75

α = 0.1  

n = 100

tα/2 = TINV(0.1,99) = 1.6604

The confidence interval for the mean is

[pic 7]           i.e.  $445.82 < μ  < $565.68

Since the bank has 1775 savings accounts thus a 90% confidence interval for the total value of all savings account balances within this bank is     1775*$445.82 < μtotal  <  1775*$565.68 i.e.,

                             $791,330.5 < μtotal  < $1,004,082

(b)

We are 90% confident that the total balance of all 1775 savings account balances within the bank are between $791,330.5 and $1,004,082.  

Problem 4

A coin operated coffee machine was designed to discharge, on the average, 7 ounces of beverage per cup.  In a test of the machine, ten cupful of coffee were drawn from the machine and measured.  The mean and standard deviation of the ten measurements were 7.2 ounces and 0.12 ounce, respectively.  Do these data present sufficient evidence to indicate that the mean discharge differs from 7 ounces?

Answer

n = 10

[pic 8]=7.2 ounce,  

s = 0.12 ounce

H0: μ = 7

HA: μ ≠ 7

α = 0.05 (we assume alpha to be 5% since the question does not tell us any value for alpha)

[pic 9]

tα/2 =  TINV(0.05,9)= 2.262  [this is a two tailed test]

Thus the rejection region is > 2.262 and <-2.262.

Since the tcalculated lies in the rejection region, thus we reject the null hypothesis. In other words our data does provide evidence that the mean discharge differs from 7 ounces

Problem 5

BatCo (The Battery Company) produces your typical consumer battery. The company claims that their batteries last at least 100 hours, on average. Your experience with the BatCo battery has been somewhat different, so you decide to conduct a test to see if the company’s claim is true. You believe that the mean life is actually less than the 100 hours BatCo claims. You decide to collect data on the average battery life (in hours) of a random sample and the information related to the hypothesis test is presented below.  

...