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# Hypothesis Testing

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[1.]
A recent article in the Topeka Times reported that 40% of all farms in Kansas will have financial problems this year. A recent survey of 200 farms revealed that 66 of them had financial problems. At the &#945; = 0.01 significance level can we conclude that a smaller proportion of farms actually have problems?

a. State the null and alternative hypothesis statements

b. See the list of formulas at the end of this test to determine the appropriate formula that would be used to calculate the test statistic for this hypothesis test.

c. The test statistic was calculated to be -2.02, determine the p-value.

d. Determine the 99% confidence interval estimate of the proportion of Kansas farms that will have financial problems.

e. Given that the Topeka Times article's estimate of 40% of the farms will have financial problems was a good estimate, how large a sample should they have taken to have been 95% confident in their estimate within an error of 5%?

f. Interpret the results of the hypothesis test, decision and summary.

[2.]
An office manager knows from long experience that 7 percent of the administrative staff will arrive late to work at least once this week.

a. What is the probability that in a sample of 15 staff members none will arrive late at least once this week? ANSWER:

b. What is the probability that in a sample of 15 staff members at least one will arrive late at least once this week? ANSWER:

[3.]
A recent study of the cost per meal in family restaurants in the downtown New York City area were normally distributed with a mean of \$18.00 and a standard deviation of \$2.50.

a. If a single cost of a meal was selected at random what is the probability that the meal would cost less than \$16.00?

b. If a sample of 5 meal costs were selected, what is the probability that the sample mean would be less than \$16.00?

[4.]
In your own words (please do not give me a quote from the textbook) write a paragraphexplaining what the p-value is.

[5.]
A large international retail chain is interviewing recent college graduates for potential positions with the company. Each graduate is rated as either below average, average or above average with respect to technical knowledge and each graduate is also rated with respect to his or her potential for advancement; fair, good, or excellent. These traits for 500 recent graduates are summarized in the following table.

Technical Knowledge Fair Good Excellent Total
Bellow average 16 12 22 50
Average 45 60 45 150
Above average 93 72 135 300
Total 154 144 202 500

a. If a single recent graduate was selected at random what is the probability they would have technical knowledge or have excellent potential for advancement?

b. If a single recent graduate was selected at random what is the probability they would have above average technical knowledge and good potential for advancement?

c. If a single recent graduate was selected at random what is the probability they would have below average sales ability given they have good potential for advancement?

[6.]
In your own words (please do not give me a quote from the textbook) write a paragraph defining "Inferential Statistics".

[7.]
A consumer organization wants to know whether there is a significant difference in the price of a particular model of portable cd players at three different types of stores. The price of the cd player was checked in a sample of 5 discount stores, 5 variety stores and 5 department stores. The results are shown below (all prices are in dollars). Test to see if there is a significant difference in the prices between the three types of stores using a significance level of &#945; = 0.05. The test statistic for this test was calculated to be 13.38.

Discount Variety Department
12 15 19
13 17 17
14 14 16
12 18 20
15 17 19

a. State the null and alternative hypothesis statements.

b. Determine the critical value.

c. See the list of formulas at the end of this test to determine the appropriate formula that would be used to calculate the test statistic for this hypothesis test.

d. Interpret the results.

[8.]
At a local pizza restaurant the owner believes that 50% of the customers will pay by cash, 25% will pay with a credit card and another 25% will pay with a personal check. The owner asked the manager to collect a random sample of two hundred customers to see their method of payment and test to see if the owner was correct, the survey results are below. Can it be
concluded that the owner is correct? Use the significance level of &#945; = 0.05.

Cash Credit Card Check
Frequency 112 48 40

a. State the null and alternative hypothesis statements.

b. Determine the critical value.

c. See the list of formulas at the end of this test to determine the appropriate formula that would be used to calculate the test statistic for this hypothesis test.

d. The test statistic was calculated to be 3.52, interpret the results.

[9.]
In your own words, write a paragraph defining what a "Hypothesis Test" is. (Don't give me an example of one, nor tell me how to construct one; tell me what it is, its use, and its interpretations.)

[10.]
A psychologist conducted a recent study to determine whether the fact that whether a student was left or right-handed was independent of their ability to perform will in a statistics class. The following are the results of the study. The test statistic for this test was calculated to be 18.18.

Handedness
Ability Left Right
Bellow average 40 52
Average 106 47
Above average 67 32

What is your conclusion using a significance level of &#945; = 0.05

a. State the null and alternative hypothesis statements.

b. Determine the critical value.

c. See the list of formulas at the end of this test to determine the appropriate formula that would be used to calculate the test statistic for this hypothesis test.

d. Interpret the results.

[11.]
The research department at the home office of New Haven Insurance conducts ongoing research on the causes of automobile accidents, the characteristics of the drivers and so on. A random sample of 400 policies written on single persons revealed 120 had at least one accident in the previous three-year period. Similarly, a sample of 600 policies written on married persons revealed that 150 had been in at least one accident. At the &#945; = 0.05 level of significance, is there a significant difference in the proportions of single and married persons having an accident during the last three-year period? For this test, the test statistic was calculated to be -1.74.

a. State the null and alternative hypothesis statements.

b. See the list of formulas at the end of this test to determine the appropriate formula that would be used to calculate the test statistic for this hypothesis test.

c. Determine the p-value.

d. Interpret the results.

[12.]
The director of the Atlanta office for Delta Airlines would like to compare the weights of checked baggage for families of four people and compare those weights between families who are traveling overseas for vacations against those families of four who are flying within the United States for vacation. She collected the following sample information:

Overseas Domestic
x = 142.50 ñb x = 130.25 lb
s = 12.25 lb s = 15.75 lb
n = 6 n = 7

The degrees of freedom for this comparison were calculated to be 10. At the &#945; = 0.10 level of significance, can she conclude that the mean weights of the checked baggage for the families flying overseas is greater than the weights of the families making domestic flights?

a. State the null and alternative hypothesis statements.

b. Determine the critical value.

c..See the list of formulas at the end of this test to determine the appropriate formula that would be used to calculate the test statistic for this hypothesis test.

d. The test statistic for this test was calculated to be 1.58, interpret the results.

[13.]
At a General Electric refrigerator assembly plant, the director of human resources has noticed an increase in number of hours lost to on-the-job accidents. Four years ago, in an attempt to improve the situation, he began a safety awareness program in which employees heard safety announcements and suggestions during their lunch hour in the cafeteria. To evaluate the program, he selected a random sample of eight participants and found the number of hours each lost in the six months before the safety awareness program began and in the last six months. Below are the results. At the &#945; = 0.05 significance level, can he conclude that the number of hours lost has declined?

Employee Before After
1 6 5
2 6 2
3 7 1
4 7 3
5 4 3
6 3 6
7 5 3
8 6 7

a. State the null and alternative hypothesis statements.

b. Determine the critical value.

c. See the list of formulas at the end of this test to determine the appropriate formula that would be used to calculate the test statistic for this hypothesis test.

d. The test statistic for this test was calculated to be -1.70, interpret the results.

https://brainmass.com/statistics/hypothesis-testing/hypothesis-testing-187668

#### Solution Preview

1. A recent article in the Topeka Times reported that 40% of all farms in Kansas will have financial problems this year. A recent survey of 200 farms revealed that 66 of them had financial problems. At the &#945; = 0.01 significance level can we conclude that a smaller proportion of farms actually have problems?

a. State the null and alternative hypothesis statements.

H0: The proportion of all farms in Kansas with financial problems is greater than or equal to 40%; p >= 0.40
H1: The proportion of all farms in Kansas with financial problems is less than 40%; p < 0.40

b. See the list of formulas at the end of this test to determine the appropriate formula that would be used to calculate the test statistic for this hypothesis test.

We will do a one-proportion z-test. The formula is:

where and p = population proportion, q = 1 - p, and n = sample size.

c. The test statistic was calculated to be -2.02, determine the p-value.
The observed value of z = -2.02 is NOT less than the critical value of z = -2.33. Therefore we cannot reject the null hypothesis.

The p-value associated with -2.02 is 0.0217 I used a calculator here: http://faculty.vassar.edu/lowry/ch6apx.html. This is significant at the 0.05 level but not at the 0.01 level.

d. Determine the 99% confidence interval estimate of the proportion of Kansas farms that will have financial problems.

p +/- z *&#8730;(p)(1-p)/n

0.33 +/- 2.575* &#8730;(0.33)(0.67)/200

0.33 +/- 2.575 * 0.0332

0.33 +/- 0.08549

e. Given that the Topeka Times article's estimate of 40% of the farms will have financial problems was a good estimate, how large a sample should they have taken to have been 95% confident in their estimate within an error of 5%?
p +/- z *&#8730;(p)(1-p)/n

0.33 +/- 1.96*&#8730;(0.33)(0.67)/n

We want 1.96*&#8730;(0.33)(0.67)/n to be less than or equal to 5%:

1.96*&#8730;(0.33)(0.67)/n <= 0.05

&#8730;(0.33)(0.67)/n <= 0.02551

(0.33)(0.67)/n <= 0.00065

1/n <= 0.00294

n >= 339.75

f. Interpret the results of the hypothesis test, decision and summary.

We cannor reject the null hypothesis that the proportion is greater than or equal to 0.40 at the 0.01 significance level. We must assume that the cited proportion of 40% was correct (the true population proportion is not less than that). The 99% confidence interval estimate is: 0.33 +/- 0.08549. In order to make a 95% confidence interval accurate to within 5%, we need a sample size of at least 340 people.

Edit in any additional work for Question #1 here: EE

2. An office manager knows from long experience that 7 percent of the administrative staff
will arrive late to work at least once this week.

a. What is the probability that in a sample of 15 staff members none will arrive late at least once this week?

This is an example of a binomial experiment.

1. There are n trials. n = 15
2. Each trial results in a success or a failure. A "success" will be being late.
3. The probability of a success, p, is constant. p = 0.07
4. The trials are independent.

The formula to calculate the probability that there will be k successes in n trials is:

P(0 out of 15 people will be late) = 15!/(0!)(15 - 0)! * (0.07)^0 * (0.93)^15

= 15!/15! * 1 * 0.3367

= 1 * 1 * 0.3367

= 0.3367 = 33.67%

b. What is the probability that in a sample of 15 staff members at least one will arrive late at least once this week?

We're looking for the probability that one or more people came in late. We can find this by subtracting the probability that no one came in late (0.3367) from one:

1 - 0.3367 = 0.6633 = 66.33%

Edit in any additional work for Question #2 here: EE

3. A recent study of the cost per meal in family restaurants in the downtown New York City area were normally distributed with a mean of \$18.00 and a standard deviation of \$2.50.

a. If a single cost of a meal was selected at random what is the probability that the
meal would cost less than \$16.00?

We will use z-scores.

z = (16.00 - 18.00)/2.50 = -0.8

Consult a z-table to see that the probability associated with this z-score is 0.212, or 21.2%.

b. If a sample of 5 meal costs were selected, what is the probability that the sample
mean would be less than \$16.00?

z = (16.00 - 18.00) / (2.5/&#8730;5) = -2/1.118 = -2.236

The probability associated with this z-score is 0.013 = 1.3%.

Edit in any additional work for Question #3 here: EE

4. In your own words (please do not give me a quote from the textbook) write a paragraph
explaining what the p-value is.

The p-value is a value ...

\$2.19