Ciency:1. The score on the chapter 8 test is compared with a student's chapter 2 test score. Theclaim that there is no difference in the test scores is to be tested.a)Find thet test statisticChapter 84042504546Chapter 235404245d.010b) Find the margin of error.c) Find the 95% confidence interval estimate of the mean difference .For a 90% confidence interval between the difference of 2 independent proportions (home runs per

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Asked Dec 5, 2019
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Ciency:
1. The score on the chapter 8 test is compared with a student's chapter 2 test score. The
claim that there is no difference in the test scores is to be tested.
a)
Find thet test statistic
Chapter 8
40
42
50
45
46
Chapter 2
35
40
42
45
d.
0
10
b) Find the margin of error.
c) Find the 95% confidence interval estimate of the mean difference .
For a 90% confidence interval between the difference of 2 independent proportions (home runs per
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Ciency: 1. The score on the chapter 8 test is compared with a student's chapter 2 test score. The claim that there is no difference in the test scores is to be tested. a) Find thet test statistic Chapter 8 40 42 50 45 46 Chapter 2 35 40 42 45 d. 0 10 b) Find the margin of error. c) Find the 95% confidence interval estimate of the mean difference . For a 90% confidence interval between the difference of 2 independent proportions (home runs per

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Expert Answer

Step 1

(a)

Step by step procedure to obtain the test statistic using MINITAB software is given below:

  • Choose Stat > Basic Statistics > 1-Sample t.
  • In One or more samples, each in a column, select Difference.
  • Check Options, enter Confidence level as 95.
  • In Perform Hypothesis test, enter 0 Under Hypothesized mean.
  • Choose Mean ≠ Hypothesized mean in alternative.
  • Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

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One-Sample T: Difference Descriptive Statistics 95% CI for N Mean StDev SE Mean 5 4.20 4.21 1.88 (-1.02, 9.42) e mean of Difference Test Null hypothesis Alternative hypothesis Heμ- 0 Hi: u = 0 T-Value P-Value 2.23 0.089

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Step 2

From the MINITAB output, the value of test statistic is 2.33.

(b)

The degrees of freedom is obtained as follows:

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df = n -1 = 5 –1 = 4

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Step 3

The critical value is obtai...

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fx =T.INV.2T(0.05,4) A1 D 2.776445

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