I need help with multiple assignments in a business statistics subject (Data 300)
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I need help with multiple assignments in a business statistics subject (Data 300). Please propose your own price.
I need help with multiple assignments in a business statistics subject (Data 300)
1 Class 3 Statistical Techniques to Test Business Hypotheses Business statistics help describe and analyze business problems, which is vital for both business professionals and everyday life. Often business owners want to know how their decisions will affect the business. According to Harvard Business School, before making decisions, y ou must first test the hypothesis of a possible solution . We will never be 100% sure that this decision is correct but by testing hypotheses, we can be more confident in our decision. A business hypothesis is an assumption about a business, sales or customers, finances, or business parameters that can be confirmed or refuted through research. At what stages of business or product development should hypotheses be tested? Modern business is developing rapidly, so it is better to test hypotheses at al l stages such as : idea, market analysis, launch, scaling, and entering new markets . An algorithm that allows you to test a hypothesis can be briefly written as a HADI (Hypothesis Action Data Insights) cycle. The cycle consists of four blocks: 1) Hypothesis – the formation of a hypothesis; 2) Action – check; 3) Data – getting the result; 4) Insights – conclusions (Fig. 1). When testing business hypotheses, we will use the methods of mathematical statistics. Fig . 1 We will use the sampling method most often used in statistics. It is information gathered from a sample taken from the general population that can be used to formulate certain judgments about the entire general population. Any statistical conclusions obtained after processing the sample are called statistical hypotheses. Statistical hypotheses about the values of the parameters of the characteristics in the general population are called parametric . And hypotheses about the law of distribution of a sign of the general population are called non -parametric . In a specific situation, a statistical hypothesis is formulated as an assumption, with a certain level of significance , about the properties of the general population based on estimates of the sample population. The boundary of the unlikely is considered = . i.e. 5% (sometimes 0.01 or 0.001). Null and Alternative Hypotheses The hypothesis to be tested is called the null hypothesis and denoted by H 0 (“H sub naught” ). An alternative hypothesis is one that contradicts the null hypothesis, it is denoted by H 1 (“H sub one” ) (Fig. 2) The alternative hypothesis, H 1 is an assertion that holds if the null hypothesis is false. The null and alternative hypotheses include all possible values of the population parameter, so either one must be false. There are three possible choices for the set of null and alternative hy potheses to be used for a given test. Described in terms of an (unknown) population mean ( ), they might be listed as shown below. Notice that each null hypothesis has an equality term in its statement (i.e.”=”, “≥”, “≤”) : Fig. 2 2 Null Hypothesis Alternative Hypothesis Description 0: = $10 1: ≠ $10 is $10 or it i ′ 0: ≥ $10 1: < $10 is at least $10 or less . 0: ≤ $10 1: > $10 is no more than $10 or it is more . Null hypothesis Alternative hypothesis ➢ Begin with the assumption that the null hypothesis is true. Similar to the notion of innocent until proven guilty. ➢ Always contains “=“, or “≤”, or “≥” sign ➢ May or may not be rejected, according to the test results . ➢ Is the opposite of the null hypothesis ➢ Contains “≠“, or “<”, or “>” sign ➢ May be supported or not supported by evidence that comes from data . A hypothesis is called simple if it is described by a single number, for example, 0: = $10 . A hypothesis is called complex if it is ambiguous, for example, 0: ≤ $10 . In this case, a complex hypothesis consists of many simple hypotheses, for example 0: = $9 ,0: = $8 ,…, 0: = $2 . Let ’s look at examples of null and alternative statistical hypotheses formulated in relation to the general population . In order to choose one b e sure to stay aware of the fact that in each case both hypotheses cannot be true at the same time . Example 1 . H0: “Average tire life is 35,000 miles.” H0: μ = 35,000 miles Н1: μ ≠ 35,000 miles Example 2 . H0: “Average tire life is at least 35,000 miles.” H0: ≥ 35 ,000 miles H1: < 35 ,000 miles Example 3 . H0: “Average tire life is no more th an 35,000 miles.” H0: ≤ 35 ,000 miles H1: > 35 ,000 m. Statistical Criterion To test the hypothesis H 0, a statistical criterion is chosen, according to which the null hypothesis is rejected or not rejected. A statistical criterion is a random variable with a probability distribution law known to us in advance. We will use two types of criteria for testing . ➢ The first test statistic (σ known ) has the standard normal distribution. = ̅− √ ⁄ ➢ The second test statistic (σ unknown ) has Student’s t-distribution with (n−1 ) degrees of freedom. = ̅− √ ⁄ , where – population mean , – standard deviation , – estimate standard deviation , − sample size . The population must be normally distributed (Fig. 3) . Fig . 3 The probability density function of the normal distribution has the following form: (,): ()= 1 √2−12(− )2 , (0,1): ()= 1 √2−2 2 (− distribution ). Student’s t -distribution has the probability density function given by : 3 Mean , Dispersion , Standard deviation and Confidence Interval Co Population Sample population Mean value of a feature = ∑ ̅= ∑ Dispersion 2= ∑(− )2 − 1 2= ∑(− ̅)2 − 1 Standard Deviation = √∑(− )2 − 1 = √∑(− ̅)2 − 1 Confidence interval ̅± ∙ √ , ≥ 30 ̅± (,−1)∙ √ , < 30 Fig. 4 Directional and Non -directional testing ➢ A directed claim is obtained when the population parameter is greater than (>), at least (≥), no more than (≤), or less than (<) some . For example, a supplier claims that no more than 20% of cans are dented. ➢ A non -directional claim is obtained when the parameter is equal to some . For example, in the statement that 35% of transit passengers are senior citizen . Directional assertions lead to one -sided (one -tail) tests , where the null hypothesis can be rejected by an extreme result in one direction. A non -directional assertion involves a two -sided (two -tail) test in which a null hypothesis can be rejected by an extreme result occurring in either direction (Fig. 5). Left -sided test Right -sided test Two -sided test : < :> : ≠ Fig. 5 For the left -sided (right -sided) tests , the rejection area consists of one part, which is located to the left (right) of the center and is determined by one critical point. In a two -sided test , the rejection area consists of two parts: one on the left and one on the right and is determined by two critical points . 4 Example 4. H0: “35% of the transit are senior citizen s”. H0: = 0.35 , H1: ≠ 0.35 , where denotes the population proportion . The alternative hypothesis H1 contains sign “≠ “,so we have a two – sided critical region (Fig. 6a). Example 5. H0: “No more than 20% of cans are dented ”. H0: ≤ 0.2, H1: > 0.2. The alternative hypothesis H1 contains sign “> “, so we have a right – sided critical region (Fig. 6b). Fig. 6 Outcomes and Type I and Type II Errors When you perform a hypothesis test, there are four possible outcomes depending on the actual validity (or falseness) of the null hypothesis H0 and the decision you have to reject it or not . The four possible outcomes in the table are as follows : 1. The decision is not to reject 0 when H 0 is true (correct decision ). 2. The decision is to reject 0 when 0 is true (incorrect decision known as a Type I error). 3. The decision is not to reject 0 when, in fact, 0 is false (incorrect decision known as a Type II error). 4. The decision is to reject 0 when 0 is false (correct decision whose probability is called “Power of the Test ”). Each of the errors occurs with a particular probability. The Greek letters α and β represent the probabilities. ➢ α = probability of a Type I error = P(Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true. ➢ β = probability of a Type II error = P(Type II error) = probability of not rejecting the null hypothesis when the null hypothesis is false. α and β should be as small as possible because they are probabilities of errors. They are rarely zero. The Power of the Test is (1 – β). Ideally, we will want a high power that is as close to one as possible. Increasing the sample size can increase the Power of the Test. HYPOTHESIS TESTING: BASIC PROCEDURES 1. Formulate the null and alternative hypotheses: 0 ,1. 2. Select the significance level ( : 0.1, 0.05, 0.01) 3. Select the test statistic z or t and calculate value , by formulas: = ̅− 0 √ ⁄ = ̅− 0 √ ⁄ 5 3. Identify Critical Values for the Test Statistic and State the Decision Rule. According to the null and alternative hypotheses, a right -sided, left -sided or two -sided critical region is constructed, namely: 0∶= 0 then if: ➢ 1: > 0, then a right -sided critical region is chosen ; ➢ 1: < 0, then the left -sided critical region is chosen ; ➢ 1: ≠ 0, then a two -sided critical region is chosen. To construct a critical region (left -sided, right -sided or two -sided), it is necessary to find critical points. In accordance with the chosen statistical criterion and significance level α, critical points are found in tables (Fig. 7 -8) or calculated in Google spreadsheets: (), (,) , = − 1− number of degrees of freedom , - sample size. () Fig. 7 Fig. 8 For calculations critical point , we will use Google Sheets (Fig. 9) , but you may also use the slider (Fig.10) if you follow the link below. https://www.mathsisfun.com/data/standard -normal -distribution -table.html Calculations in Goog le sheets Fig. 9 Fig. 10 (,) See the Google Sheets of the t - distribution (Fig.11) to find the critical point (, ). Follow the link https://www.desmos.com/calculator/fajhbwtjlc?lang=en (Fig. 1 2) 6 Calculations in Google sheets Fig. 11 Fig.12 4. Compare calculated , and critical values , and reach a conclusion about the null hypothesis. Depending on the calculated value of the test statistic, it will fall into either a rejection region or a nonrejection region. ➢ If the calculated value , is in a rejection region, the null hypothesis will be rejected. ➢ Otherwise, the null hypothesis cannot be rejected. 5. Make the related business decision. After rejecting or failing to reject the null hypothesis, the results are applied to the business decision situation . Two -Tail Testing of a Mean, Known Problem 1. When a robot welder is in adjustment, its mean time to perform its task is 1.3250 minutes. Past experience has found the standard deviation of the cycle time to be 0.0396 minutes. An incorrect mean operating time can disrupt the efficiency of other activities along the production line. For a recent random sample of 80 jo bs, the mean cycle time for the welder was 1.3229 minutes. Does the machine appear to be in need of adjustment? Solution. 1. Formulate the Null and Alternative Hypotheses 0:= 1.3250 minutes The machine is in adjustment. 1:≠ 1.3250 minutes The machine is out of adjustment. In this test, we are concerned that the machine might be running at a mean speed that is either too fast or too slow. Accordingly, the null hypothesis could be rejected by an extreme sample result in either direction , hence we have a two -tailed test . The hypothesized value for the population mean is 0= 1.3250 minutes, shown at the center of the distribution in Fig .13. Fig. 13 7 2. Select the Significance Level The significance level used will be = 0.05 . If the machine is running properly, there is only a 0.05 probability of making the mistake of concluding that it requires adjustment. 3. Select the Test Statistic and Calculate Its Value The population standard deviation () is known and the sample size is large = 80 , so the normal distribution is appropriate and the test statistic will be z, calculated as : = ̅− 0 √ ⁄ = 1.3229 − 1.3250 0.0396 /√80 = −0.0021 0.00443 = −0.47 4. Identify Critical Values for the Test Statistic and State the Decision Rule ➢ Considering 1:≠ 1.3250 , we conclude that we have a two -sided critical region. ➢ Therefore, we find two critical points that are symmetric with respect to 0. We find them according to the table of critical points for z. ➢ For a two -tail test using normal distribution and = 0.05 , = −1.96 and = +1.96 will be the respective bounda ries for lower and upper tails of 0.025 each. ➢ These are the critical values for the test, and they identify the rejection and nonrejection regions shown in Fig. 13. ➢ The decision rule can be stated as “Reject H 0 if calculated < −1.96 or > +1.96 , otherwise do not reject.” 5. Compare Calculated and Critical Values and Reach a Conclusion for the Null Hypothesis ➢ The calculated value, = −0.47 , falls within the nonrejection region of Fig. 13. ➢ At the . level of significance, the null hypothesis cannot be rejected. 6. Make the Related Business Decision ➢ Based on these results, the robot welder is not in need of adjustment. ➢ The difference between the hypothesized population mean, 0= 1.3250 minutes, and the observed sample mean, ̅= 1.3229 , is considered to have been merely the result of chance variation . Using Google Spreadsheets to Test Hypotheses Fig. 14 shows the perform ed calculations, and Fig. 15 plots the standard normal distribution f(z) that we used to test the hypothesis. Fig. 14 Fig. 15 8 Note. We can construct a 95% confidence interval for : ̅± ∙ √= 1.3229 ± 1.96 0.0396 √80 , 1.3142 1.3316 minutes . Note that the hypothetical value, 0= 1.3250 minutes, is within the 95% confidence interval, 1.3142 < . < 1.3316 meaning the confidence interval tells us it could be 1.3250 minutes. One -Tail Testing of a Mean, Known Problem 2. Light bulbs in an industrial warehouse were found to have an average lifespan of 1030.0 hours with a standard deviation of 90.0 hours. The warehouse manager was contacted by a representative of a company that produces lamps with a life extension device. The manager is concerned that the average life of new light bulbs with the device cannot exceed 1030 .0 hours . The manager decide s to test 40 light bulbs equipped with the device and f inds that their average life was 1061.6 hours. Do new bulbs really work better ? Solution . 1. Formulate the Null and Alternative Hypotheses The warehouse manager is concern ed that the bulbs might not be any better than those used in the past . This leads to a directional test. Accordingly, the null and alternative hypotheses are: 0:≤ 1030 hours New device is no better than the existing one. 1:> 1030 hours New device increases lamp life. At the center of the hypothesized distribution will be the highest possible value for which H 0 could be true, 0= 1030 hours. 2. Select the Significance Level The level chosen for the test will be = 0.05 . If this device really has no favorable effect, the maximum probability of our mistakenly concluding that will be 0.05 . 3. Select the Test Statistic and Calculate Its Value As in the previous test, the population standard deviation () is known and the sample size is large = 40 , so the normal distribution is appropriate and the test statistic will be z. It is calculated as = ̅− 0 √ ⁄ = 1061 .6− 1030 90 /√40 = 2.22 . 4. Select the Critical Value for the Test Statistic and State the Decision Rule For a right -tail z -test in which = 0.05 , = +1.645 will be the boundary separating the nonrejection and rejection regions. This critical value for the test is included in Fi g. 1 6. The decision rule can be stated as “Reject H 0 if calculated > +1.645 , otherwise do not reject.” Fig. 1 6 9 5. Compare Calculated and Critical Values and Reach a Conclusion for the Null Hypothesis The calculated value, z =2.22, falls within the rejection region of the diagram in Fig . 16. When the value is a t a 0.05 level of significance, the null hypothesis is rejected. 6. Make the right business decision The results show that the new device does increase the average lamp life. The difference between the hypothetical distribution 0= 1030 hours, and the observed sample mean, ̅= 1061 .6, is considered too large to have happened by chance. A firm may wish to incorporate lamps with a new fixture into their warehouse lighting syste m. Using Google Spreadsheets to Test Hypotheses Fig. 17 shows the performed calculations to test the null hypothesis for Problem 2. Fig. 17 Other Levels of Significance This test was conducted at the 0.05 level, but would the conclusion have been different if other levels of significance had been used instead? Consider the following possibilities: • For a significance level of 0.025. The critical value of z is 1.96, and the calculated value of z = 2.22 exceeds it. The null hy pothesis is rejected, and we again conclude that the device increases the light bulb’s life. • For a significance level of 0.005. The critical value of z is 2.58, and the calculated value of z = 2.22 does not exceed it. The null hypothesis is not rejected and we conclude that the device does not increase lamp life (Fig. 1 8). Fig . 1 8 This means that 0.5% of 100% of the lamps with the new device will no t be better than the old ones. As these possibilities suggest, using different levels of significance can lead to quite different conclusions. 10 The p -val ue Approach to Hypothesis Testing There are two basic approaches to conducting a hypothesis test: • Using a predetermined level of significance, establish critical value(s), then see whether the calculated test statistic s fall into a rejection region for the test. This is similar to placing a high -jump bar at a given height, then seeing whether you can clear it. • Determine the exact level of significance associated with the calculated value of the test statistic. In this case, we’re identifying the most extreme critical value that the test statistic would be capable of exceeding. This is equivalent to your jumping as high as you can with no bar in place, then having the judges tell you how high you would have cleared if t here had been a crossbar. There is some level of significance (p -value) at which the computed value of the test statistic exactly matches the critical value. For a given dataset, the p -value is sometimes referred to as the observed significance level. Thi s is the lowest possible level of significance at which the null hypothesis can be rejected. For the ” lamps” test, the calculated value of the test statistics was z = 2.22. For a critical z = 1.645, the area of the right tail can be found using the normal distribution in Google Sheet s. Referring to the normal distribution table, we see that 2.22 standard error units to the right of the mean includes a cumulative area of 0.9868, leaving (1.0000 – 0.9868), or 0.0132, in the right -tail area. This identifies the most demanding level of significance that ” Lamp ” could have achieved. If we had originally specified a significance level of 0.0132 for our test, the critical value for z would have been exactly the same as the value calculated. Thus, the p -value for t he ” Lamp ” test is found to be 0.0132 (Fig.1 9). The ” Lamp ” example was a one -tail test —and accordingly, the p -value was the area in just one tail (Fig. 20 ). Calculations in Google Sheets Fig. 1 9 Fig. 20 For two -tail tests, such as the “R obot Welder ” example of Figure 1 3, the p -value will be the sum of both tail areas, as shown in Fig .21. The calculated test statistic was z =0.47, resulting in a cumulative area of 0.3192 in the left t ail of the distribution. Since the “R obot Welder ” test was two -tail, the 0.3192 must be multiplied by 2 to get the p -value of 0.6384 (Fig. 22). Calculations in Google Sheets Fig. 21 Fig. 22 11 Interpreting the p -value to accept the hypothesis . − ≥ − Don ′t reject the null hypothesis . − < − Reject the null hypothesis . Problem 1. − = 0.63836 > 0.05 − 0:= 1.3250 do not rejected . Problem 2. − = 0.01321 < 0.05 − 0: ≤ 1030 is rejected . Two -Tail Testing of a Mean, Unknown in Google Sheets Problem 3. The credit manager of a large department store claims that the mean balance for the store’s charge account customers is $410. An independent auditor selects a random sample of 18 accounts and finds a mean balance of x = $511.33 and a standard deviation of s= $183.75. If the manager’s claim is not supported by the data, the auditor will examine all charge account balances. If the population of account balances is assumed to be approximately normally distributed, what action should the auditor take? Solution. 1. Formulate the Null and Alternative Hypotheses 0:= $410 The mean balance is actually $410. 0:≠ $410 The mean balance is some other value. In evaluating the manager’s claim, a two -tail test is appropriate since it is a non directional statement that could be rejected by an extreme result in either direction. The center of the hypothesized distribution of sample means for samples of = 18 will be 0= $410 . 2.Select the Significance Level . For this test, we will use the = 0.05 . 3.Select the Test Statistic and Calculate Its Value . Use the t statistic, since the population standard deviation is unknown and = $183 .75 . The sampling distribution has an estimated standard error of ̅= √= $183 .75 √18 = $43 .31 , and the calculated value of t will be = ̅− 0 ̅ = 511 .33 − 410 43 .31 = 2.34 5. Identify Critical Values for the Test Statistic and State the Decision Rule . For this test, = 0.05 , and the number of degrees of freedom will be = (− 1) = 17 . Calculate (, ) and − . Doing the calculations in a Google Sheet (Fig. 23) , we get (Fig. 24) : Fig. 2 3 Fig. 2 4 12 Results shown in Fig. 2 3-24 and the decision rule can be stated as “Reject H 0 if the calculated t is either, < −2.11 0 > 2.110 , otherwise do not reject.” In this t -test, we used the student’s t-distribution, the distribution function for the degree of freedom = 17, (problem 2) is shown in Fig. 25. Fig. 25 6. Compare the Calculated and Critical Values and Reach a Conclusion for the Null Hypothesis . The calculated test statistic, = 2.34 , exceeds the upper boundary and falls into this rejection region and so − < : 0.0318 < 0,05 . H0 is rejected. 7. Make the Related Business Decision . The results suggest that the mean charge account balance is some value other than $410. The auditor should proceed to examine all charge account balances. One -Tail Testing of a Mean, Unknown in Google Sheets Problem 4 . The Chekzar Rubber Company, in financial difficulties because of a poor reputation for product quality . They have come out with an ad campaign claiming that the mean lifetime for Chekzar tires is at least 60,000 miles for highway use. Skeptical, the editors of a consumer magazine purchase 36 of the tires and test them . The mean tire life in the sample is ̅= , . miles, with a sample standard deviation of = , . miles . Solution . 1. Formulate the Null and Alternative Hypotheses . Because of the directional nature of the ad claim and the editors’ skepticism regarding its truthfulness, the null and alternative hypotheses are 0: ≥ $ 60 ,000 miles The mean tire life is at least 60,000 miles. 1: < 60 ,000 miles The mean tire life is under 60,000 miles. 2. Select the Significance Level . For this test, the significance level will be specified as = 0.01 . 3. Sele ct the Test Statistic and Calculate Its Value The test statistic is t and the t distribution will be used to describe the sampling distribution of the mean for samples of = 36 . The center of the distribution is the lowest possible value for which H 0 could be true, or 0= 60 ,000 miles. Since the population standard deviation is unknown, s is used to estimate . The samplin g d istribution has an estimated standard error of ̅= √= 3,632 .53 miles √36 = 605 .42 miles , and the calculated value of t will be 13 = ̅− 0 ̅ = 58 ,341 − 60 ,000 .00 605 .42 = −2.739 . 4. Identify the Critical Value for the Test Statistic and State the Decision Rule . For this test, has been specified as 0.01. The number of degrees of freedom is = ( – 1), (36 − 1)= 35 . The t distribution table is now used in finding the value of t that corresponds to a one -tail area of 0.01 and df =35 degrees of freedom. We find critical value in Google Sheets (see Fig. 2 6): = −2.438 . (Although the value listed is positive, remember that the distribution is symmetrical, and we are looking for the left -tail boundary.) The rejection and nonrejection regions are shown in Fig . 27. The d ecision rule can be stated as “Reject H 0 if the calculated is less than 2.438, otherwise do not reject.” 5. Compare the Calculated and Critical Values and Reach a Conclusion for the Null Hypothesis . The calculated test statistic, t= -2.739, is less than the critical value, t = -2.438, and falls into the rejection region of the test. The null hypothesis , 0: ≥ $ 60 ,000 miles , must be rejected . When the population standard deviation is unknown Computer Solutions Fig. 2 6 shows us how we can use Google Sheets to carry out a hypothesis test for the mean . Fig. 2 6 Fig. 2 7 Using a comparison of p -value and significance level, we can conclude that: − < : 0.0048 < 0.01 . H0 is rejected. 6. Make the Related Business Decision . The test results support the editors’ doubts regarding Chekzar’s ad claim. The magazine may wish to exert either readership or legal pressure on Chekzar to modify its claim. Practical assignment Based on your data , create one problem like the example s 1 -4 in the lecture and solve it using test hypothesis in Google Sheets. Do calculations and make a plot of distribution. Report: Upload to the Canvas practical assignment for this class as 2 screenshot s: 1) calculations 2) plot End of week 3
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