Question

In: Math

group hours 1.00 44.00 1.00 40.00 1.00 44.00 1.00 39.00 1.00 25.00 1.00 37.00 1.00 31.00...

group hours
1.00 44.00
1.00 40.00
1.00 44.00
1.00 39.00
1.00 25.00
1.00 37.00
1.00 31.00
1.00 40.00
1.00 22.00
1.00 34.00
1.00 39.00
1.00 20.00
1.00 39.00
1.00 42.00
1.00 41.00
2.00 36.00
2.00 40.00
2.00 37.00
2.00 35.00
2.00 39.00
2.00 40.00
2.00 36.00
2.00 38.00
2.00 24.00
2.00 27.00
2.00 29.00
2.00 24.00
2.00 45.00
2.00 44.00
2.00 44.00
3.00 52.00
3.00 50.00
3.00 51.00
3.00 52.00
3.00 45.00
3.00 49.00
3.00 47.00
3.00 46.00
3.00 47.00
3.00 47.00
3.00 46.00
3.00 45.00
3.00 50.00
3.00 47.00
3.00 49.00

The senior class at a charter school specializing in the physical sciences surveyed students in three specialization tracks; physics, chemistry and math.

Using the data provided above. Provide the answers to the following questions starting with:

The null hypothesis for this study is u physics = u chemistry = u math

1. The alternate hypothesis is:

a. u physics =/= u math =/= u chemistry

b. All of the means are different

c. At least 2 of the means are different

d. At least two of the means are the same

2. The Value of the ANOVA statistic (F) is:

3. The probability that we would get this large a value for ANOVA is (p value):

4. Based on this result we would draw which conclusion

a. Reject the null hypothesis and accept the alternative hypothesis that at least two of the population means are different

b. Reject the null hypothesis and accept the alternative hypothesis that at least two of the sample means are different

c. Reject the null hypothesis and accept the alternative hypothesis that all of the population means are different

d. Fail to Reject the null hypothesis

Solutions

Expert Solution

using SPSS>analyze>means>one way ANOVA

ANOVA                  
hours
  Sum of Squares df   Mean Square   F   Sig.
Between Groups   1529.378   2   764.689   20.016   .000
Within Groups   1604.533   42   38.203      
Total 3133.911   44      

  

The null hypothesis for this study is u physics = u chemistry = u math

1. The alternate hypothesis is:

c. At least 2 of the means are different

2. The Value of the ANOVA statistic (F) is: 20.016

3. The probability that we would get this large a value for ANOVA is (p value): 0.000

4. Based on this result we would draw which conclusion

a. Reject the null hypothesis and accept the alternative hypothesis that at least two of the population means are different


Related Solutions

group hours 1.00 44.00 1.00 40.00 1.00 44.00 1.00 39.00 1.00 25.00 1.00 37.00 1.00 31.00...
group hours 1.00 44.00 1.00 40.00 1.00 44.00 1.00 39.00 1.00 25.00 1.00 37.00 1.00 31.00 1.00 40.00 1.00 22.00 1.00 34.00 1.00 39.00 1.00 20.00 1.00 39.00 1.00 42.00 1.00 41.00 2.00 36.00 2.00 40.00 2.00 37.00 2.00 35.00 2.00 39.00 2.00 40.00 2.00 36.00 2.00 38.00 2.00 24.00 2.00 27.00 2.00 29.00 2.00 24.00 2.00 45.00 2.00 44.00 2.00 44.00 3.00 52.00 3.00 50.00 3.00 51.00 3.00 52.00 3.00 45.00 3.00 49.00 3.00 47.00 3.00 46.00 3.00 47.00...
Calculate the PH after the addition of 0.00, 5.00, 15.00, 25.00, 40.00, 45.00,49.00, 50.00, 51.00, 55.00...
Calculate the PH after the addition of 0.00, 5.00, 15.00, 25.00, 40.00, 45.00,49.00, 50.00, 51.00, 55.00 and 60mL of 0.100M HCl in the tirtration of 50.00mL of 0.100M weak Monoprotic base B. Kb of B is 9.52 *10^-7
A mutual fund manager has a $40.00 million portfolio with a beta of 1.00. The risk-free...
A mutual fund manager has a $40.00 million portfolio with a beta of 1.00. The risk-free rate is 4.25%, and the market risk premium is 6.00%. The manager expects to receive an additional $29.50 million which she plans to invest in additional stocks. After investing the additional funds, she wants the fund's required and expected return to be 13.00%. What must the average beta of the new stocks be to achieve the target required rate of return? Do not round...
A mutual fund manager has a $40.00 million portfolio with a beta of 1.00. The risk-free...
A mutual fund manager has a $40.00 million portfolio with a beta of 1.00. The risk-free rate is 4.25%, and the market risk premium is 6.00%. The manager expects to receive an additional $29.50 million which she plans to invest in additional stocks. After investing the additional funds, she wants the fund's required and expected return to be 13.00%. What must the average beta of the new stocks be to achieve the target required rate of return? Do not round...
How many grams of 50 wt% NaOH (FM 40.00) should be diluted to 1.00 L to...
How many grams of 50 wt% NaOH (FM 40.00) should be diluted to 1.00 L to make 0.10 M NaOH? (Answer with two digits.)
A stock is currently priced at $37.00. The risk free rate is 5% per annum with...
A stock is currently priced at $37.00. The risk free rate is 5% per annum with continuous compounding. In 7 months, its price will be either $42.18 or $31.82. Using the binomial tree model, compute the price of a 7 month bear spread made of European puts with strike prices $41.00 and $45.00.
The average time spent sleeping​ (in hours) for a group of medical residents at a hospital...
The average time spent sleeping​ (in hours) for a group of medical residents at a hospital can be approximated by a normal​ distribution, as shown in the graph to the right. Answer parts​ (a) and​ (b) below. mu equals 5.4 hours sigma equals 1.2 hour A graph titled "Sleeping Times of Medical Residents" has a horizontal axis labeled "Hours" from 1 to 10 in increments of 1. A normal curve labeled mu = 5.4 hours and sigma = 1.2 hour...
The average time spent sleeping​ (in hours) for a group of medical residents at a hospital...
The average time spent sleeping​ (in hours) for a group of medical residents at a hospital can be approximated by a normal​ distribution, as shown in the graph to the right. Answer parts​ (a) and​ (b) below. mu equals 5.4 hours sigma equals 1.2 hour A graph titled "Sleeping Times of Medical Residents" has a horizontal axis labeled "Hours" from 1 to 10 in increments of 1. A normal curve labeled mu = 5.4 hours and sigma = 1.2 hour...
The length of time, in hours, it takes an "over 40" group of people to play...
The length of time, in hours, it takes an "over 40" group of people to play one soccer match is normally distributed with a mean of 2.2 hours and a standard deviation of 0.55 hours. A sample of size n = 80 is drawn randomly from the population. Find the probability that the sample mean is between 2.1 hours and 2.4 hours.
The length of time, in hours, it takes an "over 40" group of people to play...
The length of time, in hours, it takes an "over 40" group of people to play one soccer match is normally distributed with a mean of 4 hours and a standard deviation of 0.5 hours. A sample of size n = 50 is drawn randomly from the population. Find the probability that the sample mean is between 3.5 hours and 4.1 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT