In order to compare the durability of four different brands of golf balls (ALPHA, BEST, CENTURY, and DIVOT), the National Golf Association randomly selects five balls of each brand and places each ball into a machine that exerts the force produced by a 250-yard drive. The number of simulated drives needed to crack or chip each ball is recorded. The results are given in the following table. Alpha Best Century Divot
281 270 218 364
220 334 244 302
274 307 225 325
242 290 273 337
251 331 249 355
Provide the EXCEL printout of ONEWAY ANOVA using either DATA ANALYSIS or PTSTAT2 to determine whether the mean number of stimulated drives differs by brand of golf ball.
Use the ANOVA printout to write-up the six step hypothesis testing procedure for determining whether the average number of simulated drives differs by brand of golf ball. What is the p-value for this problem and what does it mean? Assume alpha =.05.

As our computed test statistic falls in critical region, so we reject our null hypothesis. Also p-value=0.0009 is less than significance level, so we reject our null hypothesis and conclude that the average number of simulated drives differs by brand of golf ball at 5% significance level.

Step-by-step explanation:

Excel output print out

Anova: Single Factor      

SUMMARY      

GroupsCountSumAverageVariance  

Alpha51268253.6609.3  

Best51532306.4740.3  

Century51209241.8469.7  

Divot51583316.61235.3  

ANOVA      

Source of VariationSSdfMSFP-valueF crit

Between Groups20960.436986.89.1492175730.0009258043.238871517

Within Groups12218.416763.65    

Total33178.819    

The six steps of hypothesis are

1. Null hypothesis: The average number of simulated drives are same by brand of golf ball

Alternative hypothesis: The average number of simulated drives differs by brand of golf ball

2. Significance level: α=0.05

3. Test statistic: F-ratio for one way anova

4. Computations

F=9.149

5. Critical region: F>3.239

6. Conclusion:

As our computed test statistic falls in critical region, so we reject our null hypothesis. Also p-value=0.0009 is less than significance level, so we reject our null hypothesis and conclude that the average number of simulated drives differs by brand of golf ball at 5% significance level.

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752