{"id":5454,"date":"2022-08-30T03:52:00","date_gmt":"2022-08-30T03:52:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/?post_type=chapter&p=5454"},"modified":"2022-10-02T12:01:25","modified_gmt":"2022-10-02T12:01:25","slug":"14a-preview","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/chapter\/14a-preview\/","title":{"raw":"14A Preview","rendered":"14A Preview"},"content":{"raw":"Preparing for the next class\r\n\r\nIn the next in-class activity, you will need to understand when to consider using a one way ANOVA and how to write the null and alternative hypotheses for a one-way\u00a0 ANOVA.\r\n\r\nIn In-Class Activities 13.C and 13.D, we explored hypothesis tests that allowed us to\u00a0 compare means from two groups\/populations. More specifically, we performed\u00a0 calculations to determine if there was evidence that the means associated with the\u00a0 populations were statistically different from one another.\r\n\r\nIn the next few in-class activities, we will learn about one-way ANOVA (analysis of\u00a0 variance), which is a statistical test for comparing and making inferences about means\u00a0 associated with two or more groups. The one-way ANOVA is also referred to as the\u00a0 one-factor ANOVA.\r\n
\r\n

Question 1<\/h3>\r\n1) Suppose a researcher wants to investigate the effect of the amount of fertilizer on\u00a0 the height of a common houseplant. More specifically, the researcher is interested in\u00a0 determining if there is a difference between the mean heights of the plants receiving\u00a0 one of three different fertilizer levels: high, medium, and low.\r\n\r\nPart A: True or False: The researcher should consider conducting a two-sample t test in this situation.\r\n\r\nPart B: True or False: The researcher should consider conducting a one-way\u00a0 ANOVA in this situation.\r\n\r\n<\/div>\r\nThe null hypothesis for a one-way ANOVA states that all the group\/population means are the same. This can be written as:\r\n\r\n[latex]H_{0}:\\mu_{1}=\\mu_{2}=\\ldots=\\mu_{k}[\/latex]\r\n\r\nwhere [latex]k[\/latex] is the number of independent groups or samples.\r\n
\r\n

Question 2<\/h3>\r\n2) Given the previous fertilizer scenario, which of the following would be the correct null hypothesis?\r\n
    \r\n \t
  1. a) \ufffd0: \ufffd1 = \ufffd2<\/li>\r\n \t
  2. b) \ufffd\": \ufffd1 = \ufffd2 = \ufffd3<\/li>\r\n \t
  3. c) \ufffd0: \ufffd1 = \ufffd2 = \ufffd3 = \ufffd4<\/li>\r\n \t
  4. d) \ufffd+: \ufffd1 = \ufffd2 = \ufffd3<\/li>\r\n<\/ol>\r\n<\/div>\r\nThe alternative hypothesis for a one-way ANOVA is a bit different than the alternative\u00a0 hypothesis we used when comparing only two group means (i.e., two-sample t-test).\r\n\r\nWhen there were only two group means to consider, the null hypothesis that the two\u00a0 means were the same was [latex]H_{0}:\\mu_{1}=\\mu_{2}[\/latex]. If you wanted to show that the two means were\u00a0 different or not equal, the alternative hypothesis would be [latex]H_{A}:\\mu_{1}\\neq\\mu_{2}[\/latex]. If we rejected the null hypothesis, we would be able to conclude that the two means were statistically\u00a0 different.\r\n\r\nWhen we reject the null hypothesis for a one-way ANOVA, we cannot simply state that\u00a0 all of the means are not equal. That is, when we reject the null hypothesis, [latex]H_{0}:\\mu_{1}=\\mu_{2}=\\ldots=\\mu_{k}[\/latex], we are not able to differentiate whether one of the means is\u00a0 different from the others, whether two of the means are different from the others,\u00a0 whether three of the means are different from the others, etc.\r\n\r\nSo, to provide flexibility and to account for the multiple outcomes associated with\u00a0 rejecting the null hypothesis, the alternative hypothesis for a one-way ANOVA should be\u00a0 written as:\r\n\r\n[latex]H_{A}:[\/latex] At least two of the group means are different.\r\n
    \r\n

    Question 3<\/h3>\r\n3) Given the previous fertilizer scenario, which of the following would be the correct\u00a0 alternative hypothesis?\r\n
      \r\n \t
    1. a) \ufffd+: \ufffd1 \u2260 \ufffd2<\/li>\r\n \t
    2. b) \ufffd+: \ufffd1 \u2260 \ufffd2 \u2260 \ufffd3<\/li>\r\n \t
    3. c) \ufffd+: All of the group means are different.<\/li>\r\n \t
    4. d) \ufffd+: At least two of the group means are different.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
      \r\n

      Question 4<\/h3>\r\n4) A researcher would like to determine whether there is a difference between the\u00a0 means of exercise hours per week among people in the U.S. regions of the\u00a0 Northeast, South, West, and Midwest.\r\n\r\nWhich of the following best explains why a one-way ANOVA should be considered\u00a0 for this situation?\r\n\r\na) The researcher is interested in comparing the means of two groups.\r\n\r\nb) The researcher is interested in comparing the means of three groups.\r\n\r\nc) The researcher is interested in comparing the means of four groups.\r\n\r\n<\/div>\r\n
      \r\n

      Question 5<\/h3>\r\n5) Which of the following is the correct null hypothesis for the scenario described in\u00a0 Question 4?\r\n
        \r\n \t
      1. a) \ufffd0: \ufffd1 = \ufffd2<\/li>\r\n \t
      2. b) \ufffd\": \ufffd1 = \ufffd2 = \ufffd3<\/li>\r\n \t
      3. c) \ufffd0: \ufffd1 = \ufffd2 = \ufffd3 = \ufffd4<\/li>\r\n \t
      4. d) \ufffd0: \ufffd1 \u2260 \ufffd2 \u2260 \ufffd3 \u2260 \ufffd4<\/li>\r\n<\/ol>\r\n<\/div>\r\n
        \r\n

        Question 6<\/h3>\r\n6) Which of the following is the correct alternative hypothesis for the scenario described\u00a0 in Question 4?\r\n
          \r\n \t
        1. a) \ufffd+: At least two of the group means are different.<\/li>\r\n \t
        2. b) \ufffd+: At least three of the group means are different.<\/li>\r\n \t
        3. c) \ufffd+: All of the group means are different.<\/li>\r\n \t
        4. d) \ufffd+: \ufffd% \u2260 \ufffd' \u2260 \ufffd. \u2260 \ufffd\/<\/li>\r\n<\/ol>\r\n<\/div>\r\n
          \r\n

          Question 7<\/h3>\r\n7) A researcher is interested in conducting a one-way ANOVA to compare means\u00a0 between five groups. Which of the following would be the null and alternative\u00a0 hypotheses for this situation?\r\n
            \r\n \t
          1. a) \ufffd\": \ufffd% = \ufffd' = \ufffd. = \ufffd\/ = \ufffd0<\/li>\r\n<\/ol>\r\n\ufffd+: The group means are different.\r\n
              \r\n \t
            1. b) \ufffd0: \ufffd1 = \ufffd2 = \ufffd3 = \ufffd4 = \ufffd0<\/li>\r\n<\/ol>\r\n\ufffd+: At least three of the group means are different.\r\n
                \r\n \t
              1. c) \ufffd\": \ufffd% = \ufffd' = \ufffd. = \ufffd\/ = \ufffd0<\/li>\r\n<\/ol>\r\n\ufffd+: \ufffd% \u2260 \ufffd' \u2260 \ufffd. \u2260 \ufffd\/ \u2260 \ufffd0\r\n
                  \r\n \t
                1. d) \ufffd0: \ufffd1 = \ufffd2 = \ufffd3 = \ufffd4 = \ufffd0<\/li>\r\n<\/ol>\r\n\ufffd+: At least two of the group means are different.\r\n\r\n<\/div>","rendered":"

                  Preparing for the next class<\/p>\n

                  In the next in-class activity, you will need to understand when to consider using a one way ANOVA and how to write the null and alternative hypotheses for a one-way\u00a0 ANOVA.<\/p>\n

                  In In-Class Activities 13.C and 13.D, we explored hypothesis tests that allowed us to\u00a0 compare means from two groups\/populations. More specifically, we performed\u00a0 calculations to determine if there was evidence that the means associated with the\u00a0 populations were statistically different from one another.<\/p>\n

                  In the next few in-class activities, we will learn about one-way ANOVA (analysis of\u00a0 variance), which is a statistical test for comparing and making inferences about means\u00a0 associated with two or more groups. The one-way ANOVA is also referred to as the\u00a0 one-factor ANOVA.<\/p>\n

                  \n

                  Question 1<\/h3>\n

                  1) Suppose a researcher wants to investigate the effect of the amount of fertilizer on\u00a0 the height of a common houseplant. More specifically, the researcher is interested in\u00a0 determining if there is a difference between the mean heights of the plants receiving\u00a0 one of three different fertilizer levels: high, medium, and low.<\/p>\n

                  Part A: True or False: The researcher should consider conducting a two-sample t test in this situation.<\/p>\n

                  Part B: True or False: The researcher should consider conducting a one-way\u00a0 ANOVA in this situation.<\/p>\n<\/div>\n

                  The null hypothesis for a one-way ANOVA states that all the group\/population means are the same. This can be written as:<\/p>\n

                  [latex]H_{0}:\\mu_{1}=\\mu_{2}=\\ldots=\\mu_{k}[\/latex]<\/p>\n

                  where [latex]k[\/latex] is the number of independent groups or samples.<\/p>\n

                  \n

                  Question 2<\/h3>\n

                  2) Given the previous fertilizer scenario, which of the following would be the correct null hypothesis?<\/p>\n

                    \n
                  1. a) \ufffd0: \ufffd1 = \ufffd2<\/li>\n
                  2. b) \ufffd”: \ufffd1 = \ufffd2 = \ufffd3<\/li>\n
                  3. c) \ufffd0: \ufffd1 = \ufffd2 = \ufffd3 = \ufffd4<\/li>\n
                  4. d) \ufffd+: \ufffd1 = \ufffd2 = \ufffd3<\/li>\n<\/ol>\n<\/div>\n

                    The alternative hypothesis for a one-way ANOVA is a bit different than the alternative\u00a0 hypothesis we used when comparing only two group means (i.e., two-sample t-test).<\/p>\n

                    When there were only two group means to consider, the null hypothesis that the two\u00a0 means were the same was [latex]H_{0}:\\mu_{1}=\\mu_{2}[\/latex]. If you wanted to show that the two means were\u00a0 different or not equal, the alternative hypothesis would be [latex]H_{A}:\\mu_{1}\\neq\\mu_{2}[\/latex]. If we rejected the null hypothesis, we would be able to conclude that the two means were statistically\u00a0 different.<\/p>\n

                    When we reject the null hypothesis for a one-way ANOVA, we cannot simply state that\u00a0 all of the means are not equal. That is, when we reject the null hypothesis, [latex]H_{0}:\\mu_{1}=\\mu_{2}=\\ldots=\\mu_{k}[\/latex], we are not able to differentiate whether one of the means is\u00a0 different from the others, whether two of the means are different from the others,\u00a0 whether three of the means are different from the others, etc.<\/p>\n

                    So, to provide flexibility and to account for the multiple outcomes associated with\u00a0 rejecting the null hypothesis, the alternative hypothesis for a one-way ANOVA should be\u00a0 written as:<\/p>\n

                    [latex]H_{A}:[\/latex] At least two of the group means are different.<\/p>\n

                    \n

                    Question 3<\/h3>\n

                    3) Given the previous fertilizer scenario, which of the following would be the correct\u00a0 alternative hypothesis?<\/p>\n

                      \n
                    1. a) \ufffd+: \ufffd1 \u2260 \ufffd2<\/li>\n
                    2. b) \ufffd+: \ufffd1 \u2260 \ufffd2 \u2260 \ufffd3<\/li>\n
                    3. c) \ufffd+: All of the group means are different.<\/li>\n
                    4. d) \ufffd+: At least two of the group means are different.<\/li>\n<\/ol>\n<\/div>\n
                      \n

                      Question 4<\/h3>\n

                      4) A researcher would like to determine whether there is a difference between the\u00a0 means of exercise hours per week among people in the U.S. regions of the\u00a0 Northeast, South, West, and Midwest.<\/p>\n

                      Which of the following best explains why a one-way ANOVA should be considered\u00a0 for this situation?<\/p>\n

                      a) The researcher is interested in comparing the means of two groups.<\/p>\n

                      b) The researcher is interested in comparing the means of three groups.<\/p>\n

                      c) The researcher is interested in comparing the means of four groups.<\/p>\n<\/div>\n

                      \n

                      Question 5<\/h3>\n

                      5) Which of the following is the correct null hypothesis for the scenario described in\u00a0 Question 4?<\/p>\n

                        \n
                      1. a) \ufffd0: \ufffd1 = \ufffd2<\/li>\n
                      2. b) \ufffd”: \ufffd1 = \ufffd2 = \ufffd3<\/li>\n
                      3. c) \ufffd0: \ufffd1 = \ufffd2 = \ufffd3 = \ufffd4<\/li>\n
                      4. d) \ufffd0: \ufffd1 \u2260 \ufffd2 \u2260 \ufffd3 \u2260 \ufffd4<\/li>\n<\/ol>\n<\/div>\n
                        \n

                        Question 6<\/h3>\n

                        6) Which of the following is the correct alternative hypothesis for the scenario described\u00a0 in Question 4?<\/p>\n

                          \n
                        1. a) \ufffd+: At least two of the group means are different.<\/li>\n
                        2. b) \ufffd+: At least three of the group means are different.<\/li>\n
                        3. c) \ufffd+: All of the group means are different.<\/li>\n
                        4. d) \ufffd+: \ufffd% \u2260 \ufffd’ \u2260 \ufffd. \u2260 \ufffd\/<\/li>\n<\/ol>\n<\/div>\n
                          \n

                          Question 7<\/h3>\n

                          7) A researcher is interested in conducting a one-way ANOVA to compare means\u00a0 between five groups. Which of the following would be the null and alternative\u00a0 hypotheses for this situation?<\/p>\n

                            \n
                          1. a) \ufffd”: \ufffd% = \ufffd’ = \ufffd. = \ufffd\/ = \ufffd0<\/li>\n<\/ol>\n

                            \ufffd+: The group means are different.<\/p>\n

                              \n
                            1. b) \ufffd0: \ufffd1 = \ufffd2 = \ufffd3 = \ufffd4 = \ufffd0<\/li>\n<\/ol>\n

                              \ufffd+: At least three of the group means are different.<\/p>\n

                                \n
                              1. c) \ufffd”: \ufffd% = \ufffd’ = \ufffd. = \ufffd\/ = \ufffd0<\/li>\n<\/ol>\n

                                \ufffd+: \ufffd% \u2260 \ufffd’ \u2260 \ufffd. \u2260 \ufffd\/ \u2260 \ufffd0<\/p>\n

                                  \n
                                1. d) \ufffd0: \ufffd1 = \ufffd2 = \ufffd3 = \ufffd4 = \ufffd0<\/li>\n<\/ol>\n

                                  \ufffd+: At least two of the group means are different.<\/p>\n<\/div>\n","protected":false},"author":23592,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5454","chapter","type-chapter","status-publish","hentry"],"part":5448,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5454","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/users\/23592"}],"version-history":[{"count":3,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5454\/revisions"}],"predecessor-version":[{"id":5592,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5454\/revisions\/5592"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/parts\/5448"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapters\/5454\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/media?parent=5454"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/pressbooks\/v2\/chapter-type?post=5454"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/contributor?post=5454"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lumen-danacenter-statsmockup\/wp-json\/wp\/v2\/license?post=5454"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}