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An experiment was designed to test whether students’ learning is affected by background sound. Twenty-four students were randomly separated into three groups of eight. All students study a passage of text for 30 minutes.

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Group 1 study with music playing at a constant volume in the background. Group 2 study with recorded conversations playing in the background and group 3 study with no sound at all. After studying, all students take a multiple-choice test on the material.

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Their scores are given below:

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 Group 1 Group 2 Group 3 {r1[0]} {r2[0]} {r3[0]} {r1[1]} {r2[1]} {r3[1]} {r1[2]} {r2[2]} {r3[2]} {r1[3]} {r2[3]} {r3[3]} {r1[4]} {r2[4]} {r3[4]} {r1[5]} {r2[5]} {r3[5]} {r1[6]} {r2[6]} {r3[6]} {r1[7]} {r2[7]} {r3[7]}
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Test the hypothesis that the students' average marks are unaffected by the background sounds.

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In this example we are comparing the means of three distict samples, we therefore must use the analysis of variance test (ANOVA)

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\$$H_0:\$$ Students' average marks are unaffected by different background sounds.

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\$$H_1:\$$ Students' average marks are affected by different background sounds.

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 Group 1 Group 2 Group 3 \$${n_i}\$$ 8 8 8 \$$\\sum{x_i}\$$ {sum_r1} {sum_r2} {sum_r3} \$$\\sum{x_i^2}\$$ {r1sqd} {r2sqd} {r3sqd}
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We need to evaluate the sum of squares between the samples \$$S_b\$$, and the sum of squares within the samples \$$S_w\$$.

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\$$S_b=\\sum\\left(\\frac{(\\sum{x_i})^2}{n_i}\\right)-\\frac{(\\sum{x_{i,j}})^2}{n_t}\$$ where \$${n_i}\$$ is the number of data values in sample \$$i\$$ and \$${n_t}\$$ equals the total number of data values.

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\$$S_b=\\frac{(\\var{sum_r1})^2}{8}+\\frac{(\\var{sum_r2})^2}{8}+\\frac{(\\var{sum_r3})^2}{8}-\\frac{(\\var{total_sum_r})^2}{24}\$$

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\$$S_b=\\var{ss_betw}\$$

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\$$S_w=\\sum\\left(\\sum{x_i^2}-\\frac{(\\sum{x_i})^2}{n_i}\\right)\$$

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\$$S_w=\\left(\\var{r1sqd}-\\frac{(\\var{sum_r1})^2}{8}\\right)+\\left(\\var{r2sqd}-\\frac{(\\var{sum_r2})^2}{8}\\right)+\\left(\\var{r3sqd}-\\frac{(\\var{sum_r3})^2}{8}\\right)\$$

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\$$S_w=\\var{ss_within}\$$

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 Sum of Squares deg. of freedom Mean square F Between \$$\\var{ss_betw}\$$ 2 \$$\\var{msb}\$$ \$$\\var{test_statistic}\$$ Within \$$\\var{ss_within}\$$ 21 \$$\\var{msw}\$$
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The F-table values will have \$$F_{2,21}=2, 21\$$ degrees of freedom.

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The lower cut-off point is found using \$$\\frac{1}{F_{21,2}}\$$

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 Significance 0.05 Lower limit 0.025 Upper limit 4.42
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Compare the test statistic with the F-table values and choose your conclusion.

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Fill in the appropriate values in the table below, correct to one decimal place

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 Sum of squares degrees of freedom mean square F \nBetween\n [[0]] 2 [[2]] [[4]] \nWithin\n [[1]] 21 [[3]]
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Correct to 2 decimal places, enter the lower table statistic: F = [[5]]

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Correct to 2 decimal places, enter the upper table statistic: F = [[6]]

Having compared your test statistic with the upper and lower table values for a two-tailed F-test having 2, 21 degrees of freedom, select one of the following conclusions that best describes your conclusion.

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The test statistic falls in the upper critical region. Reject the Null Hypothesis at the 5% level and conclude that student's average marks are affected by the background sounds.

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Accept the Null Hypothesis at the 5% significance level and conclude that student's average marks are unaffected by the background sounds.

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The test statistic falls in the lower critical region. Reject the Null Hypothesis at the 5% level and conclude that student's average marks are affected by the background sounds.

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