1 answer

Suppose a teacher recorded the attendance of her students in a recent statistics class because she...

Question:

Suppose a teacher recorded the attendance of her students in a recent statistics class because she wanted to investigate the linear relationship between the number of classes they missed and their final grades. The accompanying table shows these data for a random sample of nine students

Classes Missed Final Grade
4   73
6   81
1   92
4   72
0   94
2   86
0   89
5   87
2   96 Suppose a teacher recorded the attendance of her students in a recent statistics class because she wanted to investigate the

Suppose a teacher recorded the attendance of her students in a recent statistics class because she wanted to investigate the linear relationship between the number of classes they missed and their final grades. The accompanying table shows these data for a random sample of nine students. Complete parts a through c. Click the icon to view the table showing missed classes and final grade. a. Calculate the correlation coefficient for this sample. The correlation coefficient is (Type an integer or decimal rounded to three decimal places as needed.) b. Using a = 0.05, test to determine if the population correlation coefficient is less than zero. The hypotheses for this test are The test statistic is (Type an integer or decimal rounded to two decimal places as needed.) The p-value is (Type an integer or decimal rounded to three decimal places as needed.) sufficient evidence to conclude that Therefore, at the 0.05 level of significance, the population correlation coefficient is the null hypothesis. There zero. C. onclusions can be on ese results? The conclusion is that there linear relationship between missed classes and final grade.
Math Statistics And Probability
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a) The correlation coefficient is given by

r={n\sum xy -(\sum x)(\sum y)\over \sqrt{[n\sum x^2-(\sum x)^2][n\sum y^2-(\sum y)^2]}}

Take x as represents number of missed classes and y denotes final grades

using the data given we get

r\approx -0.639

b) The null and the alternative hypothesis are

H_0: \rho =0 \\~~~~~~ H_a: \rho<0

The value of the test static is t={r\sqrt{n-2}\over \sqrt{1-r^2}}={-0.639\sqrt{9-2}\over \sqrt{1-(-0.639)^2}}\approx -2.20

The test is left tailed test, number of degrees freedom is 9-2=7

So P-value P(t<-2.20)=0.032

Since 0.032<0.05, we reject the null hypothesis.

There is sufficient evidence to support the claim that correlation coefficient is less than zero

c) There is negative correlation between the two

.

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