Question

In: Statistics and Probability

Conduct an Independent Samples T test to answer the questions based on the following scenario. (Assume...

Conduct an Independent Samples T test to answer the questions based on the following scenario. (Assume a non-directional research hypothesis (two-tailed test) and a level of significance of .05) The superintendent who collected data for Assignments 1 and 2, continued to examine the district’s data. One question that concerned the superintendent’s constituencies was the difference between the school performance scores of the superintendent’s district and a neighboring district that had similar demographic and socio-economic characteristics. The superintendent collected the following information: School performance scores for superintendent’s district: 62 66 73 79 64 68 75 85 69 63 60 71 51 78 65 79 58 86 76 63 60 80 45 61 72 63 59 School performance scores for comparison district: 67 68 69 67 73 66 64 60 61 61 56 60 57 72 62 63 64 70 65 63 67 63 57 60 64 69 68

1. What are the mean and standard deviation for the superintendent’s district?

2. What are the mean and standard deviation for the comparison district?

3. State an appropriate null hypothesis for this analysis.

4. What is the observed or computed value of t?

5. What is the value of the degrees of freedom that are reported in the output (equal variances assumed)?

6. What is the reported level of significance?

7. Based on the reported level of significance, would you reject the null hypothesis?

8. Present the results as they might appear in an article. This must include a table and narrative statement that reports and interprets the results of the Independent Samples T test. Note: The table must be created using your word processing program. Tables that are copied and pasted from SPSS are not acceptable.

Solutions

Expert Solution

  1. The mean of superintendent's district() is calculated as:

    The standard deviation of superintendent's () district is:



  2. The mean of comparison district() is calculated as:

    The standard deviation of comparison () district is:

  3. An appropriate null hypothesis will be:
    vs

  4. The test statistic is given by:

    where,





  5. The values of degrees of freedom is (27+27-2)=52(since, there are 27 samples of each, and two means are estimated)

  6. The reported level of significance is


  7. We reject the null hypothesis, if

    Howener, in this case,

    Hence, the null is not rejected.

  8. Mean Standard Deviation T_statistic
    Superintendent 67.81481 9.957745
    1.672817
    Comparison 64.2963 4.504825

    From the above table, it is observed that there exists no difference between the school performance scores of the superintendent’s district and a neighboring district that had similar demographic and socio-economic characteristics. Both the districts are performing equally with relatively low variance.

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