Homework Assignment # 4 Problem MBS-2. Rush Hour Traffic in ActivStats (Tests for a Mean Homework) In order to reduce the average number of vehicles that use the Lincoln Tunnel, the Port Authority in New York is experimenting a peak-hour pricing, . The mean number of cars waiting in a queue is 1220 (population mean: m = 1220).
A random sample of 10 days (n = 10) is drawn and the data representing the number of cars is analyzed.
The Data Desk tool in ActivStats reports that the sample mean is 989. 8 cars (sample mean: y = 989. 8) and the sample standard deviation is 160. 676 cars (s = 160. 676).
In order to determine whether the peak-hour pricing has contributed for reducing the number of cars using the tunnel, we have to assume that the numerical data are independently drawn and represent a random sample from a population that is normally distributed.
Therefore, as the sampling distribution of the mean is normally distributed and the population standard deviation (s) is not known, it is appropriate to use the t-test. The null and the alternative hypothesis for this test are: Ho: m = 1220 (or less) and Ha: m > 1220. For this problem we can use 1% level of significance (a = 0. 01).
Because s is not known, we chose a t-test with a test statistic t, given by the formula: t = (y – m) / set = (y – m) / (s / √ n) This test is one-sided and a critical value (t ) is needed for identifying the value of the test statistic that is required to reject the null hypothesis. t (df = n-1, a) t (df = 10-1, a = 0.
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01) t (df = 9, a = 0. 01) The critical value t , which I obtained from the t-distribution table in Kazmier, and corresponds to 9 degrees of freedom and 0. 01 level of significance is t = 2. 821 On the basis of the collected data, we can compute the t-test statistic: t = (y – m) / (s / √ n) t = (989. 8 – 1220) / (160. 676/√ 10) (after replacing for y, m, s and n) t = – 4.
5306 Therefore, in order to reject the null hypothesis the sample mean must have a value that is bigger than the critical value (reject Ho if t > t , otherwise do not reject Ho).
Because t = -4, 5306 falls within the non-rejection region (Fig. 1) below the critical value t = 2. 821, we cannot reject Ho.
We can conclude that there is no evidence that the average number of cars has changed from the originally observed amount of 1220, thus the difference that we observed is insignificant and subject to change. Using the Data Desk tool, we can obtain the p-value which is p = 0. 007 (Fig. 2).
Therefore, the probability of obtaining a test statistic equal to or more extreme than the result obtained from the sample data is equal to 0. 007, given that the null hypothesis Ho is really true.
Problem MCS-2. Effective Repellent in ActivStats (Tests for a Mean Homework) An experiment is conducted to test the reliability of a new mosquito repellent. The summary statistics obtained from a random sample of 5 people (n = 5) are: sample mean y = 83% (y = 0. 83) and sample standard deviation s = 15% (s = 0. 15).
1.
As the population standard deviation (σ ) is not known, it is appropriate to use a t – test. A “one-sided test” is to be conducted, with a null and alternative hypothesis as follows: Ho: μ = 95% = 0. 95 (or more) Ha: μ < 0. 95 For this test we will use a 10% level of significance (α = 0. 1).
The critical value (t ) co responds to: t (df = n-1, a) t (df = 5-1, a = 0.
1) t (df = 4, a = 0. 1) t = 1. 533 (as obtained from the t-distribution table in Kazmier) On the basis of the collected data we calculate the value of the test statistic t according to the following formula: t = (y – μ ) / (s / √ n) t = (0. 83 – 0. 95) / (0. 15 / √ 5) t ~ – 1.
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789 If the null hypothesis is to be rejected, we have to obtain a value of the sample mean that is bigger than the critical value. Reject Ho if t > t Otherwise do not reject Ho As illustrated in Figure 3, the test statistic t = -1. 789 falls in the non-rejection area (above t = 1. 533).
Therefore, there is no evidence that proves the new repellent to be inefficient.
P value? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 2. The assumptions we have to make in order to have a valid hypothesis test are regarding the distribution of the population. Thus, we assume that the numerical data in this sample are independently drawn and represent a random sample from a population that is normally distributed. Problem MRB-4. Will They Charge More? in ActivStats (Tests for a Mean Homework) A bank experiments a special offer and omits the annual credit card fee for 200 of its customers (n = 200) that charge at least $2400 (μ = 2400) in a year. For this particular sample, the mean increase of the amount the customers charge is $332, therefore = 2400+332 y = 2732 $ 1.
The significance level for this test is 1% (α = 0. 01) and the null and alternative hypothesis are: Ho: μ = 2400 (or more) Ha: μ < 2400 A t-test is to be conducted, for which we need to compare the test statistic t with the critical value t . The critical value t can be obtained form the t-distribution table in Kazmier. t (df = n-1, a) t (df = 200-1, a = 0. 01) t (df = 199, a = 0. 01) t = 2.
326 As the t-distribution approaches the normal, and the number of observations exceed 120 (n = 200), we can assume that s is approximately equal to σ . The test statistic can be obtained through the following formula: t = (y – m) / set = (2732 – 2400) / 108 (replacing for y, m, and sx) t = 3. 0740 Reject Ho if t > t Otherwise do not reject Ho After comparing the values for the test statistic t and t , we can reject the null hypothesis, as the sample mean has a value bigger than the critical value (Fig. 5).
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Introduction to Hypothesis Testing 8.1 Inferential Statistics and Hypothesis Testing LEARNING OBJECTIVES 8.2 Four Steps to Hypothesis Testing After reading this chapter, you should be able to: 8.3 Hypothesis Testing and Sampling Distributions 8.4 Making a Decision: Types of Error 8.5 Testing a Research Hypothesis: Examples Using the z Test 8.6 Research in Focus: Directional Versus Nondirectional ...
Therefore, we can conclude that there is no evidence that proves an increase in the amount charged under the no-fee offer. 2. If a large enough number of samples of the same size n = 200 are taken and their sample means are computed, 99% of them will include the true population mean somewhere within the interval around their sample means, and only 1% of them would not. The 99% confidence interval for the mean can be calculated in the following way: lower bound = y – t df, α /2 (s/ √ n) = 2732 – 2.
5758 (108/ √ 200) = 2732 – 19. 6707 = 2712. 3293 upper bound = y + t df, α /2 (s/ √ n) = 2732 + 2. 5758 (108/ √ 200) = 2732 + 19.
6707 = 2751. 6707 d [ 2712. 3293 < y < 2751. 6707 ]3. As the number of observations in this sample is large enough, n = 200, we can apply the Central Limit Theorem. This, with a large enough number of observations, the sampling distribution of the mean can be approximated by the normal distribution.
This statement holds true, regardless of the shape of the individual values in the population. 4.
