Understanding Basic Statistics
Understanding Basic Statistics
8th Edition
ISBN: 9781337558075
Author: Charles Henry Brase, Corrinne Pellillo Brase
Publisher: Cengage Learning
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Chapter 11, Problem 15CR

(a)

To determine

To test: The verification of Se=0.518.

(a)

Expert Solution
Check Mark

Answer to Problem 15CR

Solution: It is verified that, Se0.647.

Explanation of Solution

Calculation: Let x represent the batting average of a professional baseball player and y represent the home run percentage.

We have, x=1957,y=30.1,x2=0.553,y2=150.15,

xy=8.753.

We have to calculate first a,b as follows:

b=n(xy)(x)(y)nx2(x)2=7(8.753)(1.957)(30.1)7(0.553)(1.957)255.166

x¯=xn=1.95770.2796

y¯=yn=30.174.3

a=y¯bx¯=4.355.166(0.2796)11.1244

Se=y2aybxyn2=150.15+11.1244(30.1)55.166(8.753)72=0.647450.647

Conclusion: Hence, it is verified that, Se0.647

(b)

To determine

To test: The verification of r0.948 and the claim that ρ>0 at 0.01 level of significance.

(b)

Expert Solution
Check Mark

Answer to Problem 15CR

Solution: It is verified that r0.948. We have sufficient evidence to conclude that the population correlation coefficient between x and y is positive at 1% level of significance.

Explanation of Solution

Calculation:

The correlation coefficient, r is calculated as follows:

r=n(xy)(x)(y)[nx2(x)2][ny2(y)2]=7(8.753)(1.957)(30.1)[7(0.553)(1.957)2][7(150.15)(30.1)2]0.948

Using the level of significance, α=0.01.

The null hypothesis for testing is defined as,

H0:ρ=0

The alternative hypothesis is defined as,

H1:ρ>0

The sample test statistic is,

t=rn21r2=0.948721(0.948)26.66

The degrees of freedom are d.f.=n2=72=5

The above test is right tailed test, so we can use the one-tail area in the student’s distribution table (Table 4 of the Appendix). From table, the p-value for the sample test statistic t=6.66 is 0.0005<P-value<0.005. Since the interval containing the P-value is less than 0.01, hence we have to reject the null hypothesis at α=0.01. Therefore, we can conclude that the data is statistically significant at 0.01 level of significance.

Conclusion: It is verified that r0.948. We have sufficient evidence to conclude that the population correlation coefficient between x and y is positive at 1% level of significance.

(c)

To determine

To test: The verification of b55.166 and the claim that β>0 at 0.01 level of significance.

(c)

Expert Solution
Check Mark

Answer to Problem 15CR

Solution: It is verified that, b55.166. We have sufficient evidence to conclude that the slope is positive at 1% level of significance.

Explanation of Solution

Calculation:

The value b is calculated as follows:

b=n(xy)(x)(y)nx2(x)2=7(8.753)(1.957)(30.1)7(0.553)(1.957)255.166

Using the level of significance, α=0.01.

The null hypothesis for testing is defined as,

H0:β=0

The alternative hypothesis is defined as,

H1:β>0

The find t statistic and P-value using MINITAB software is as:

Step 1: Enter x and y in Minitab worksheet.

Step 2: Go to Stat > Regression > Regression > Fit Regression Model.

Step 2: Select ‘y’ in Response and ‘x’ in ‘Continuous predictors’ box. Then click on OK.

The sample test statistic is t=6.67 and P-value is 0.001.

The software gives the P-value for two-tailed test, for finding the p-value for one tailed test we can divide the obtained P-value by 2.

Pvalue for one tailed test=0.0012=0.0005.

Since P-value is less than 0.01, hence we can reject the null hypothesis at α=0.01. Therefore, we can conclude that the data is statistically significant at 0.01 level of significance.

Conclusion: Hence, it is verified that, b55.166. We have sufficient evidence to conclude that the slope is positive at 1% level of significance.

(d)

To determine

To test: The verification of y^11.123+55.166x and the 90% confidence interval for the predicted home run percentage for a player with a batting average of 0.310.

(d)

Expert Solution
Check Mark

Answer to Problem 15CR

Solution: It is verified that, y^11.123+55.166x. The 90% confidence interval for the predicted home run percentage for a player with a batting average of 0.310 is from 4.49 to 7.46.

Explanation of Solution

Calculation: The results obtained in above part are a11.12,b55.166

The regression line equation is y^=a+bx,

y^11.12+55.166x

The find 90% confidence interval for y when x=0.310 by using MINITAB software is as:

Step 1: Go to Stat > Regression > Regression > Predict.

Step 2: Select ‘y’ in Response and write 0.310 in ‘x’ box.

Step 3: Click on Options write 90 in ‘Confidence level’ and select ‘Two-sided’ in Type of interval. Then click on OK.

The 90% confidence interval is obtained as: (4.49,7.46).

Conclusion: Hence, it is verified that, y^0.856+0.402x. The 90% confidence interval for y when x=10 is (4.49,7.46).

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Chapter 11 Solutions

Understanding Basic Statistics

Ch. 11.1 - For Problems 9-19, please provide the following...Ch. 11.1 - For Problems 9-19. please provide the following...Ch. 11.1 - For Problems 9-19. please provide the following...Ch. 11.1 - For Problems 9-19. please provide the following...Ch. 11.1 - For Problems 9-19. please provide the following...Ch. 11.1 - For Problems 9-19. please provide the following...Ch. 11.1 - Prob. 17PCh. 11.1 - Prob. 18PCh. 11.1 - Prob. 19PCh. 11.2 - Statistical Literacy For a chi-square...Ch. 11.2 - Statistical Literacy How are expected frequencies...Ch. 11.2 - Statistical Literacy Explain why goodness-of-fit...Ch. 11.2 - Critical Thinking When the sample evidence is...Ch. 11.2 - For Problems 5-14, please provide the following...Ch. 11.2 - For Problems 5-14, please provide the following...Ch. 11.2 - For Problems 5-14, please provide the following...Ch. 11.2 - For Problems 5-14, please provide the following...Ch. 11.2 - For Problems 5-14, please provide the following...Ch. 11.2 - Prob. 10PCh. 11.2 - Prob. 11PCh. 11.2 - For Problems 5-14, please provide the following...Ch. 11.2 - Prob. 13PCh. 11.2 - Prob. 14PCh. 11.2 - Prob. 15PCh. 11.2 - For Problems 5-14, please provide the following...Ch. 11.3 - Statistical Literacy Docs the x distribution need...Ch. 11.3 - Prob. 2PCh. 11.3 - Prob. 3PCh. 11.3 - For Problems 3-11, please provide the following...Ch. 11.3 - For Problems 3-11. please provide the following...Ch. 11.3 - For Problems 3-11. please provide the following...Ch. 11.3 - Prob. 7PCh. 11.3 - For Problems 3-11, please provide the following...Ch. 11.3 - For Problems 3-11. please provide the following...Ch. 11.3 - Prob. 10PCh. 11.3 - Prob. 11PCh. 11.4 - Prob. 1PCh. 11.4 - Statistical Literacy What is the symbol used for...Ch. 11.4 - Prob. 3PCh. 11.4 - Statistical Literacy How does the t value for the...Ch. 11.4 - Prob. 5PCh. 11.4 - Using Computer Printouts Problems 5 and 6 use the...Ch. 11.4 - In Problems 7-12, parts (a) and (b) relate to...Ch. 11.4 - In Problems 7-12, parts (a) and (b) relate to...Ch. 11.4 - In Problems 7-12, parts (a) and (b) relate to...Ch. 11.4 - In Problems 7-12, parts (a) and (b) relate to...Ch. 11.4 - Prob. 11PCh. 11.4 - Prob. 12PCh. 11.4 - Prob. 13PCh. 11.4 - Prob. 14PCh. 11.4 - Prob. 15PCh. 11.4 - Prob. 16PCh. 11.4 - Prob. 17PCh. 11.4 - Prob. 18PCh. 11.4 - Prob. 19PCh. 11 - Terminology Match each of the following tests to...Ch. 11 - Prob. 2CRCh. 11 - Statistical Literacy of the following random...Ch. 11 - Prob. 4CRCh. 11 - Prob. 5CRCh. 11 - Before you solve Problems 6-10, first classify the...Ch. 11 - Prob. 7CRCh. 11 - Before you solve Problems 6-10, first classify the...Ch. 11 - Before you solve Problems 6-10, first classify the...Ch. 11 - Prob. 10CRCh. 11 - Prob. 11CRCh. 11 - Prob. 12CRCh. 11 - Prob. 13CRCh. 11 - Prob. 14CRCh. 11 - Prob. 15CRCh. 11 - The Statistical Abstract of the United States...Ch. 11 - Prob. 1LCWPCh. 11 - Prob. 2LCWPCh. 11 - Prob. 3LCWPCh. 11 - Prob. 4LCWPCh. 11 - Prob. 1CRPCh. 11 - Prob. 2CRPCh. 11 - Prob. 3CRPCh. 11 - Prob. 4CRPCh. 11 - Prob. 5CRPCh. 11 - Prob. 6CRPCh. 11 - Prob. 7CRP
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