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Statistics with Applications. 1. A random sample of 37 first graders who participated in sports had manual dexterity scores w

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Answer:

Given that:

Let Y_1 denote the manual dexterity scores of second graderas participated

Let Y_2   denote the manual dexterity scores of second graderas did not participated in sports had manual dexerity scores

Let \bar{Y_1} be the sample mean of second graders participated in sports

Let \bar{Y_2} be the sample mean of second graders did not participated in sports

Let s_1 be the standard deviation of second graders participated in sports

Let s_2   be the standard deviation of second graders participated in sports

Let n_1 denote number of second graders participated in sports

Let n_2 denote number of second graders participated in sports

The given information is as shown below

Second graders who participated in sports

The number of graders n_1=37

Sample mean \bar{Y_1}=32.19

Sample variance s_1=4.34

Second graders who did not participated in sports

The number of graders n_2=37

Sample mean \bar{Y_2}=31.68

Sample variance s_2=4.56

a)

The objective of the study is to test and determine whether there is sufficient evidence to support that second graders who participate in sports have a higher mean dextenity score

Let \mu _1 denote the population mean dexterity score for second graders who participated in sports

Let \mu _2 denote the population mean dexterity score for second graders who did not participated in sports

The null and alternative hypothesis are

H_0: \mu _1=\mu _2

VS

H_a: \mu _1>\mu _2

Use the level of significance = 0,05

From the standard normal distribution tables the critical value at 0.05 level for one tailed is 1.645

Rejection region since the test is based on right tailed reject the null hypothesis if Z>Z_a(1.645)

The test statistic is

Z=\frac{\bar{Y_1}-\bar{Y_2}}{\sqrt{\frac{s_1^2}{n}+\frac{s_2^2}{n}}}

Z=\frac{32.19-31.68}{\sqrt{\frac{4.34^2}{37}+\frac{4.56^2}{37}}}

Z=\frac{0.51}{1.0349}

Z=0.4927

Here,it can be observed that the test statistic value 0.4927 is less than the critical value 1.645.So the null hypothesis is fails to be rejected

Therefore it can be concluded that there is no sufficient evidence to support that the second graders who participate in sports have a higher dexterity score

b) Find \beta when \mu _1-\mu -2=3

From part(a) the rejection region is Z>1.645

Consider Z\geq 1.645

Z=\frac{\bar{Y_1}-\bar{Y_2}}{\sqrt{\frac{s_1^2}{n}+\frac{s_2^2}{n}}}\geq 1.645

\bar{Y_1}-\bar{Y_2}\geq 1.645{\sqrt{\frac{4.34^2}{37}+\frac{4.56^2}{37}}}

\bar{Y_1}-\bar{Y_2}\geq 1.702

Implies don't to reject H_0 if \bar{Y_1}-\bar{Y_2}<1.702

\beta =P\begin{Bmatrix} Don't\, \, reject\, \, H_o\, \, when\, \, H_1\, \, is \, \, true \end{Bmatrix}

=P(\bar{Y_1}-\bar{Y_2}<1.702\mid \mu _1-\mu _2=3)

=P(\frac{\bar{Y_1}-\bar{Y_2}-(\mu _1-\mu _2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n}}}<\frac{1.702-3}{\sqrt{\frac{4.34^2}{37}+\frac{4.56^2}{37}}})

=P(Z\leq -1.25)

=0.1056

Therefore the value of \beta when \mu _1-\mu _2=3 is 0.1056

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