A study by Hewitt Associates showed that 79% of companies offer employees flexible scheduling. Suppose a researcher believes that in accounting firms this figure is lower. The researcher randomly selects 415 accounting firms and through interviews determines that 303 of these firms have flexible scheduling. With a 1% level of significance, does the test show enough evidence to conclude that a significatly lower proportion of accounting firms offer employees flexible scheduling.
A previous experience shows the variance of a given process to be 14. Researchers are testing to determine whether this value has changed. They gather the following dozen measurements of the process. Use these data and a significance level of .05 to test the null hypothesis about the variance. Assume the measurements are normally distributed.
52 44 51 58 48 49
38 49 50 42 55 51
Highway engineers in Ohio are painting white stripes on a highway. The stripes are supposed to be approximately 10 feet long. However, because of the machine, the operator, and the motion of the vehicle carrying the equipment, considerable variation occurs among the stripe lengths. Engineers claim that the variance of stripes is not more than 16 inches. Use the sample lengths given here from 12 measured stripes to test the variance claim. Assume stripe length is normally distributed. Let the significance level = .05
Stripe Lengths in Feet
10.3 9.4 9.8 10.1
9.2 10.4 10.7 9.9
9.3 9.8 10.5 10.4.
This is one tailed t test.
The degree of freedom is 415-1=414, at 0.01 significance level, the critical value is -2.335.
The test value t=(0.73-0.79)/(sqrt(0.79*(1-0.79)/415)=-3.001
The solution discusses hypothesis testing using samples in the statistic problem set.
Problem Set: Hypothesis Test, Test Statistics, Two-Sample t-tests
1. Identify the null hypothesis and the alternative hypothesis.
A researcher claims that 62% of voters favor gun control.
2. Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis.
alpha = 0.05 for a left-tailed test.
3. Use the given information to find the P-value.
The test statistic in a right-tailed test is z = 1.43.
4. Find the P-value for the indicated hypothesis test.
In a sample of 88 children selected randomly from one town, it is found that 8 of them suffer from asthma. Find the P-value for a test of the claim that the proportion of all children in the town suffer from asthma is equal to 11%.
5. Assume that a simple random sample has been selected from a normally distributed population. Calculate the test statistic.
A researcher is testing the claim that the mean age of the prison population in one city is less than 26 years. Sample data are summarized as n = 25, x-bar = 24.4 years, and s = 9.2 years. The significance level is alpha = 0.05.
6. Decide which sampling distribution applies to the following information.
Claim: population mean, mu = 111. Sample data: n = 10, x-bar = 101, s = 15.3. The sample data appear to come from a normally distributed population with unknown mean and unknown standard deviation.
7. Formulate the indicated conclusion in non technical terms. Be sure to address the original claim.
A researcher claims that the amounts of acetaminophen in a certain brand of cold tablets have a standard deviation different from the population standard deviation, sigma = 3.3 mg, claimed by the manufacturer. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is failure to reject the null hypothesis, state the conclusion in nontechnical terms.
8. Calculate the test statistic for the following hypothesis test.
The claim is that the proportion of drowning deaths of children attributable to beaches is more than 0.25, and the sample statistics include n = 622 drowning deaths of children with 30% of them attributable to beaches.
9. Assume that a hypothesis test of the given claim will be conducted. Identify the type I error for the test.
The manufacturer of a refrigerator system for beer kegs produces refrigerators that are supposed to maintain a true mean temperature, mu, of 46 degrees Fahrenheit, ideal for a certain type of German pilsner. The owner of the brewery does not agree with the refrigerator manufacturer, and claims that he can prove that the true mean temperature is incorrect.
10. Find the critical value of Chi-Square based on the given information.
The alternative hypothesis is sigma is less than 0.629, the sample size is n = 19, and alpha is 0.025.
11. From the sample statistics, find the value of p-bar used to test the hypothesis that the population proportions are equal.
n1 = 40, x1 = 5; n2 = 445, x2 = 268.
12. Calculate the test statistic used to test the null hypothesis that p1 = p2.
A report on the nightly news broadcast stated that 15 out of 150 households with pet dogs were burglarized and 25 out of 204 households without pet dogs were burglarized.
13. Calculate the test statistic used to test the hull hypothesis that p1 = p2.
In a vote on the Clean Water bill, 41% of the 205 Democrats voted for the bill while 40% of the 230 Republicans voted for it.
14. Suppose you wish to test the claim that mu sub-d, the mean of the differences d for a population of paired data, is greater than 0. Given a sample of n = 15 and a significance level of alpha = 0.01, what criterion would be used for rejecting the null hypothesis?
15. The two data sets are dependent. Find d-bar to the nearest tenth.
X 8.5 5.9 9.1 6.7 6.8 5.7
Y 8.6 8.9 8.9 8.9 8.1 9.3
16. Two studies are described below. Decide whether they represent independent samples or matched pairs.
I. The effectiveness of a headache medicine is tested by measuring the intensity of a headache in patients before and after drug treatment. The data consist of before and after intensities for each patient.
II. The effect of caffeine as an ingredient is tested with a sample of regular soda and another sample with decaffeinated soda.
17. The differences between two sets of dependent data are 0.42, 0.42, 0.51, 0.54, and 0.6. Calculate the standard deviation of the differences, s sub-d. (Points: 5)
18. Assume that you want to test the claim that the paired sample data come from a population for which the mean difference is mu sub-d = 0. Compute the value of the t test statistic.
A farmer has decided to use a new additive to grow his crops. He divided his farm into 10 plots and kept records of the corn yield (in bushels) before and after using the additive. The results are show below.
Plot 1 2 3 4 5 6 7 8 9 10
Before 9 9 8 7 6 8 5 9 10 11
After 10 9 9 8 7 10 6 10 10 12
You wish to test the following hypothesis at the 10 percent level of significance.
H0: mu sub-d = 0 against H1: mu sub-d not = 0.
What is the appropriate test statistic?
19. When performing a hypothesis test for the ratio of two population variances, the upper critical F value is denoted F sub-R. The lower critical F value, F sub-L, can be found as follows: interchange the degrees of freedom, and take the reciprocal of the resulting F value found in table A-5. F sub-R can be denoted as F sub-alpha half and F sub-L can be denoted F sub 1-alpha half.
Find the critical values F sub-L and F sub-R for a two-tailed hypothesis test based on the following values: n1 = 9, n2 = 7, alpha = 0.05.
20. Use the computer display and Figure 7-7 on page 380 to solve the problem.
When testing for a difference between the means of a treatment group and a placebo group, the computer display below is obtained. The claim is that the treatment group (variable 1) comes from a population with a mean that is less than the mean for the placebo population. The 0.05 significance level is used. Which of the following statements is the conclusion?
t-Test: Two Samples for Means
1 Variable 1 Variable 2
2 Mean 65.10738 66.18251
3 Known Variance 8.102938 10.27387
4 Observations 50 50
5 Hypothesized Mean Difference 0
6 t -1.773417
7 P(T<=t) one-tail 0.0384
8 T Critical one-tail 1.644853
9 P(T<=t) two-tail 0.0768
10 t Critical two-tail 1.959961
There is sufficient evidence to warrant rejection of the claim that the population mean for the treatment group is significantly less than the population mean for the placebo group.
There is not sufficient evidence to warrant rejection of the claim that the population mean for the treatment group is significantly less than the population mean for the placebo group.
The sample data support the claim that the population mean for the treatment group is significantly less than the population mean for the placebo group
There is not sufficient sample evidence to support the claim that the population mean for the treatment group is significantly less than the population mean for the placebo group.