BUS-122 Applied Statistics Assignment
Instructions:
Answer all the questions on an A4 size paper, all answers must be in HANDWRITTEN form.
Attach the Cover sheet on top of the A4 answer sheets.
All staples must only be stapled on the top left hand corner.
Answer the questions in the number order as given and label them clearly.
1. If a real estate market is strong there will be a close relationship between the asking price for homes and the selling price. Suppose that one analyst believes that the mean difference between asking price and selling price for homes in a particular market area is less than $2,000. To test this using an alpha level equal to 0.05, a random sample of n = 15 homes that have sold recently was selected. The sample differences between asking price and selling price are in the following table:
Based on these sample data, what is the critical value expressed in dollars? (10 marks)
ANSWER:
The analyst claims that the mean difference will be less than $2,000. Given this research hypothesis, we can form the following null and alternative hypotheses to test statistically:
The test is a one-tail, lower-tail test with 0.05 area in the rejection region. Since we do not know the population standard deviation and the sample size is relatively small, if we assume that the population is normally distributed, the t-distribution can be used. The critical t-value for degrees of freedom 15-1 = 14 and 0.05 in one tail is –1.7613. We find the critical value as follows:
= .
We compute the sample standard deviation, s, from the data as follows: .
Then the critical value is
Thus, if < $1,769.62 reject the null hypothesis; otherwise do not reject. We find as follows: . Since $1,831.067 $1,769.62, we do not reject the null hypothesis.
2. A major U.S. oil company has developed two blends of gasoline. Managers are interested in determining whether a difference in mean gasoline mileage will be obtained from using the two blends. As part of their study, they have decided to run a test using the Chevrolet Impala automobile with automatic transmissions. They selected a random sample of 100 Impalas using Blend 1 and another 100 Impalas using Blend 2. Each car was first emptied of all the gasoline in its tank and then filled with the designated blend of the new gasoline. The car was then driven 200 miles on a specified route involving both city and highway roads. The cars were then filled and the actual miles per gallon were recorded. The following summary data were recorded:
Blend 1Blend 2
Sample Size 100 100
Sample Mean23.4 mpg 25.7 mpg
Sample St. Dev.4.0 mpg 4.2 mpg
Based on the sample data, using a 0.05 level of significance, what conclusion should the company reach about whether the population mean mpg is the same or different for the two blends? Use the test statistic approach to test the null hypothesis. (8 marks)
ANSWER:
The hypothesis test involves a test for the difference between population means and is based on large samples from each population. The appropriate null and alternative hypotheses are:
.
This will be a two-tailed test since we are interested in testing whether there is a difference between the two population means with respect to miles per gallon and neither blend of gasoline is predicted to be superior to the other.
The critical value for a two-tailed test with a significance level of 0.05 is found in the Standard Normal table to be z = +1.96. The test statistic in this case is computed using:
.
Since z = -3.9655 < -1.96, we reject the null hypothesis and conclude that the population means are not equal. Based on the samples, we infer that Blend 2 will provide higher mean mileage.
3. A PC company uses two suppliers for rechargeable batteries for their notebook computers. Two factors are important quality features of the batteries: mean use time and variation. It is desirable that the mean use time be high and the variability be low. Recently, the PC maker conducted a test on batteries from the two suppliers. In the test, 9 randomly selected batteries from Supplier 1 were tested and 12 randomly selected batteries from Supplier 2 were tested. The following results were observed:
Supplier 1Supplier 2
n1 = 9 n2 = 12
min. min.
s1 = 11.2 min s2 = 9.9 min.
Based on these sample results, can the PC maker conclude that a difference exists between the two batteries with respect to the population mean use time? Test using a 0.10 level of significance. (12 marks)
ANSWER:
This test calls for us to test whether the population means are equal. Since the sample sizes are small and the population variances are unknown, the t-distribution should be employed if we assume that the populations are normally distributed and the populations have equal variances.
We can test the following null and alternative hypotheses:
The critical value for a t-test for a two-tailed test is found from the t-distribution with degrees of freedom equal to n1 + n2 – 2 = 9 + 12 – 2 = 19. Given a 0.10 significance level, the critical value from the t-distribution with degrees of freedom equal to 19 is equal to +1.7291.
The test statistic is computed using:
where sp is the pooled standard deviation and is computed as:
.
Then the test statistic is:
.
Since t = -1.1158 > –1.7291, we do not reject the null hypothesis. Thus, based on the sample data, we do not have evidence to indicate that the two batteries differ in mean use time.
Moderate
Keywords: t-test, mean difference, independent
4. The National Football League (NFL) is interested in testing to see whether there is a difference in the proportion of male fans that prefer instant replay to review officials’ calls and the proportion of female fans that prefer instant replay. It is believed that males tend to favor the practice to a higher degree than do females. To test this, random samples of male fans and female fans were selected and the following results were obtained:
Male FansFemale Fans
Sample Size 300 100
Number Preferring234 57
Using a significance level equal to 0.05, what conclusion should be reached based on the sample data? (10 marks)
ANSWER:
The parameter of interest in this situation is the difference in population proportions. The research hypothesis states that the proportion of males favoring instant replay exceeds the proportion of females who favor it. Thus, we can formulate the following null and alternative hypotheses:
.
This means that we are dealing with a one-tailed hypothesis test. The critical value for a one-tailed test for the difference between population proportions with a 0.05 significance level is z = 1.645 from the standard normal distribution. The test statistic is computed using:
where and
and .
Then, the test statistic is:
.
Since 4.0846 > 1.645, we reject the null hypothesis and conclude that based on the sample data, the male fans do favor instant replay by a higher proportion than do female fans.
5. The Department of Weights and Measures in a southern state has the responsibility for making sure that all commercial weighing and measuring devices are working properly. For example, when a gasoline pump indicates that 1 gallon has been pumped, it is expected that 1 gallon of gasoline will actually have been pumped. The problem is that there is variation in the filling process. The state’s standards call for the mean amount of gasoline to be 1.0 gallon with a standard deviation not to exceed 0.010 gallons. Recently, the department came to a gasoline station and filled 10 cans until the pump read 1.0 gallon. They then measured precisely the amount of gasoline in each can. The following data were recorded:
Based on these data what should the Department of Weights and Measures conclude if they wish to test whether the standard deviation exceeds 0.010 gallons or not, using a 0.05 level of significance? (8 marks)
ANSWER:
The variation requirement specifies that the standard deviation can not exceed 0.01 gallon, we need to use a test for a single population variance. The appropriate null and alternative hypotheses are:
.
Note that we have converted from standard deviation to variances since there is no test for standard deviation directly. In this situation, the test statistic is a chi-square value computed as follows:
== 135.49.
This test statistic can be compared to the chi-square value from the table with 10 - 1 = 9 degrees of freedom and a one-tail area of 0.05. That value is 16.9190. Since , the null hypothesis is rejected. This means that the variation standard is being exceeded based on these data. The gasoline station will need to get the pump checked out and the variation of fill around the mean reduced.
6. A PC company uses two suppliers for rechargeable batteries for their notebook computers. Two factors are important quality features of the batteries: mean use time and variation. It is desirable that the mean use time be high and the variability be low. Recently, the PC maker conducted a test on batteries from the two suppliers. In the test, 9 randomly selected batteries from Supplier 1 were tested and 12 randomly selected batteries from Supplier 2 were tested. The following results were observed:
Supplier 1Supplier 2
n1 = 9 n2 = 12
min. min.
s1 = 11.2 min s2 = 9.9 min.
Based on these sample results, can the PC maker conclude that a difference exists between the two batteries with respect to the population standard deviations? Test using a 0.10 level of significance. (12 marks)
ANSWER:
Ideally, we would like to test whether the two population standard deviations are equal. However, there is no test available for doing this test directly. Instead, we must convert to variances and perform the F-test. The following null and alternative hypotheses would be appropriate in this situation:
This will be a two-tailed test since we are testing to see whether a difference exists and are not predicting which battery will have more or less variability. In a two-tailed test of variances, the test statistic is an F-value that is formed by the ratio of the two sample variances and by placing the larger sample variance in the numerator:
.
We then compare this value to a critical value from the F-distribution table. Since this is a twotail test with alpha = 0.10, we use the table with 0.05 in the upper-tail of the F-distribution.
Two sets of degrees of freedom are used with the F-distribution. Across the top of the F-table we look for degrees of freedom corresponding to n-1 where n is the sample size associated with the sample variance that was placed in the numerator of the F-test statistic. In our case, n = 9, so the degrees of freedom is 9-1 = 8 for the numerator. The degrees of freedom down the side in the F-table is n-1 where n corresponds to the sample size for the sample variance in the denominator of the F-test statistic. In our case, that would be n = 12. Thus, the denominator degrees of freedom are 12-1 = 11. Then the critical F from the 0.05 table with 8 and 11 degrees of freedom is 2.948. The decision rule is:
If F calculated > 2.948 reject Ho, otherwise do not reject.
Since our calculated F-test statistic is F = 1.2799 < 2.948, we do not reject the null hypothesis. Thus, based on the sample information, we have no basis for believing that there is a difference in the two batteries with respect to variability in use time.