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The amount of money a family of four spends on milk in a given year is normally
distributed with a mean of $623 and a standard deviation of $147.
a. What percentage of families of four will spend less than $500 on milk in a given year?
b. What percentage of of families of four will spend between $475 and $675 on milk in a
given year?
c. A milk company wants to give coupons out to families who are in the 99th percentile
in terms of spending for milk in a given year. What value/amount spent represents the
99th percentile?
I would use Minitab or an Excel template. I will show you both, first with Minitab
Part a
Go to Graph >> Probability Distribution Plot
Click the View Probability Box (far right), then click OK
Make sure distribution dropdown is set to Normal
Input Mean of 623 and Standard Deviation of 147
Click the Shaded Area Tab
Choose Radial Button Next to “X Value”
Click “Left Tail” since it is a “less than” situation
Input x value of 500, click OK
The graph shows us the probability is 0.2014 (our answer)
Part b
Go to Graph >> Probability Distribution Plot
Click the View Probability Box (far right), then click OK
Make sure distribution dropdown is set to Normal
Input Mean of 623 and Standard Deviation of 147
Click the Shaded Area Tab
Choose Radial Button Next to “X Value”
Click “Middle” since it is a “between two values” situation
Input x values of 475 and 675 for x value 1 and x value 2, click OK
The graph shows us the probability is 0.4812 (our answer)
Part c
Go to Graph >> Probability Distribution Plot
Click the View Probability Box (far right), then click OK
Make sure distribution dropdown is set to Normal
Input Mean of 623 and Standard Deviation of 147
Click the Shaded Area Tab
Choose Radial Button Next to “Probability” (We are finding a percentile, inputting
it as the probability)
Click “Left Tail” since it is a “Percentile” situation 99th percentile means 99 percent to
the left
Input Probability of 0.99, click OK
The graph shows us the value on the x axis (bottom right) is 965 (our answer)
That tells us that 99% of the population of families of four spend less than $965 per year
on milk (and 1% spends more)
An aerospace company produces springs for space applications. Testing is done to
determine the number of hours a spring will last when subjected to its maximum stress. A
random sample of 40 springs yields the following results.
Sample Size = 40
Sample Mean = 37.2 hours
Sample Standard Deviation = 8.5 hours
a. Compute the 90% confidence interval for the average number of hours the springs last
being tested under these conditions
b. Interpret this interval.
c. How many springs should be sampled if we want to construct a 99% confidence
interval for the number of hours a spring will last to within 3 hours?
I will use Minitab
Part a
Stat >> Basic Statistics >> 1-Sample Z (Test and Confidence Interval)
Click Radial Button next to Summarized Data
Input Sample Size of 40, Mean of 37.2 and Standard Deviation of 8.5
Click the Options button and input 90 for your Confidence level
Click Ok on both boxes
We get the following in the Session Window
One-Sample Z
The assumed standard deviation = 8.5
N
40
Mean
37.20
SE Mean
1.34
90% CI
(34.99, 39.41)
Our answer is (34.99,39.41)
Part b
We are 90% confident that the true mean time the springs will last under maximum stress
for the spring population is between 34.99 and 39.41 hours.
Part c
In Minitab
Stat >> Power and Sample Size >> Sample Size for Estimation
THE MEAN DOES NOT ENTER INTO THIS AT ALL
Enter Standard deviation of 8.5 and Margin of error for confidence intervals 3 (because it
said “within 3 hours”)
Click Options button and enter 99 because we want to be 99% confident
YOU MUST CHECK THE BOX NEXT TO “Assume population standard
deviation is known”
Click your OK buttons on both boxes
In the session window, you get your answer which is 54
I have come up with a new design for soda cans better than the pop-top design presently
used. I want to estimate the percentage of the market that this new design might capture.
In a survey, a random sample of 200 potential regular soda drinkers were asked whether
they would purchase sodas with this new design. The results were that 42 responded
affirmatively.
a. Compute the 95% confidence interval for the population proportion of potential
customers that would purchase my new design.
b. Interpret this confidence interval.
c. How many potential customers should be sampled in order to be 90% confident of
being within 1% of the population proportion of potential customers that would purchase
the new product?
The SmartDog electronic dog collar company thinks that is the best on the market. The
Marketing and Sales Vice Presidents have noted that a new campaign will not be needed
if more than 30% of the population of households with dogs prefer their design over its
competitors. The company’s market research department has distributed the new product
to 500 randomly selected households. The results of this sampling are that 145 of the 500
households prefer the SmartDog collar. Does the sample data provide evidence to
conclude that the percentage of households with dogs in the population that prefer the
SmartDog collar is more than 30% (with a = .05)? Use the hypothesis testing procedure
outlined below.
a. Formulate the null and alternative hypotheses.
b. State the level of significance.
c. Find the critical value (or values), and clearly show the rejection and nonrejection
regions.
d. Compute the test statistic.
e. Decide whether you can reject Ho and accept Ha or not.
f. Explain and interpret your conclusion in part e. What does this mean?
g. Determine the observed p-value for the hypothesis test and interpret this value. What
does this mean?
h. Does this sample data provide evidence (with a = 0.05) that the percentage of
households with dogs in the population that prefer the SmartDog collar is more than
30%?
At a data entry company, the average number of data entry mistakes per day per data
analyst was 11. The owner of the data entry company developed an auditing procedure to
decrease the number of errors. After an extended amount of time using the new auditing
procedure, the owner hired a third party who took a random sample of 50 data analysts on
randomly selected days and found the following results.
Sample Size = 50 data analysts
Sample Mean = 10.25 mistakes per day per analyst
Sample Standard Deviation = 2.65 mistakes per day per analyst
Does the sample data provide evidence to conclude that the population mean number of
data entry mistakes per day per data analyst was less than 11 (using a = .01)? Use the
hypothesis testing procedure outlined below.
a. Formulate the null and alternative hypotheses.
b. State the level of significance.
c. Find the critical value (or values), and clearly show the rejection and nonrejection
regions.
d. Compute the test statistic.
e. Decide whether you can reject Ho and accept Ha or not.
f. Explain and interpret your conclusion in part e. What does this mean?
g. Determine the observed p-value for the hypothesis test and interpret this value. What
does this mean?
h. Does this sample data provide evidence (with a = .01) that the population mean
number of register mistakes per day per clerk was less than 11?
These are just some ideas, your best bet is to go back through the homework and be
ready… You can do great!!
Best to you.