Stat 200 Final Practice Test

STAT 200 FINAL PRACTICE TEST (SOLUTIONS)

D) (note means “and”)

3A.

CI for variance or standard deviation use:

Confidence interval for population variance: Use , get two values from the table and solve for twice. (Take square root if you want )

df = 3 and the two table numbers are .216 and 9.348.

and

1337.18 to 57870.4

3B. Sample size needed for CI for mean use : sample size needed for CI for mean is

3C. Sample size needed for CI for proportion use: sample size for CI for proportion is (use to guarantee sample size is large enough, use in place of to get reasonable estimate)

3D. Difference of three means use ANOVA:

ANOVA: Collect data from SRS’s from different groups. Assume the populations are normal and the data are collected independently. Then the -statistic for rejecting: All population means are equal in favor of: There is some difference in the population means is found as follows (the sums are over each of the S sources):

where and

and

is the mean of all the data, is how pieces of data from source i, is the sample mean of all the data from source i, and is the sample variance of all the data from source i. The Anova Test is always a one-tail test to the right.

Company 1 / Company 2 / Company 3
Number of bulbs studied / 4 / 3 / 3
Lifetimes / 990,1010,900,880 / 1000,900,1200 / 850,800,1000
/ 3780 / 3100 / 2650
/ 3584600 / 3250000 / 2362500
Sample mean / 945 / 1033.33 / 883.33
Sample variance / 4166.67 / 23333.33 / 10833.33

: All population means are equal in favor

: There is some difference in the population means

If there is no difference the chance we would find such strong or stronger evidence than we got that there is a difference is over 5%. This is assuming all conditions were met and the data were obtained in a proper fashion.

3E1) Difference of two means from independent samples use:

Difference of means from 2 independent samples / t, df = min of the sample sizes - 1 / the difference of the two sample means subtracted in appropriate order / difference of the population means in Ho (often 0) /

Picture of how would be distributed if (using t with df = 9)

If there is the mean of the other company is not higher, the chance we would find such strong or stronger evidence than we got that it is higher is between 5% and 10%. This is assuming all conditions were met and the data were obtained in a proper fashion.

3E2) or

Picture of how would be distributed if (using t with df = 9)

If there is the mean of the other company is the same, the chance we would find such strong or stronger evidence than we got that it is different is between 10% and 20%. This is assuming all conditions were met and the data were obtained in a proper fashion.

3F) Comparing two variances use: Hypothesis test for ratio of two variances: Use maker sure the top is bigger than the bottom. Keep track of the two different df’s. Use the F-table for the critical value(s). Note that making F(data)>1 even if you have two critical values as in a two-tail test, only the right hand one matters.

Data summary:

Before = B / After = A
/ 240 / 233
/ 14402 / 13583
/ .6667 / 3.5833

or (note the A went on the top since A’s sample standard deviation is bigger)

Picture of how would be distributed assuming . Note the tail is .05. The df for the top is 3 and for the bottom is 3.

If the insert does not raise the variance, the chance we would find such strong or stronger evidence than we got that it does is over 5%. This is assuming all conditions were met and the data were obtained in a proper fashion.

G) Sample size needed for CI for proportion use: sample size for CI for proportion is (use to guarantee sample size is large enough, use in place of to get reasonable estimate)

H1)Comparing two means from matched pairs use:

Difference of means from 2 dependent samples (a.k.a. matched pairs) / t, df = n-1 / sample mean of the differences subtracted in appropriate order / population difference mean in Ho (often 0) / Where s is the s.d. of the differences.
Bulb 1 / Bulb 2 / Bulb 3 / Bulb 4
Power consumption before / 60 / 61 / 59 / 60
Power consumption after / 57 / 58 / 57 / 61
after - before / 3 / 3 / 2 / -1

Picture of how would be distributed if (using t with df = 3).

If the insert does not change power consumption, the chance we would find such strong or stronger evidence than we got that it does is between 10% and 20%. This is assuming all conditions were met and the data were obtained in a proper fashion.

H2)

Picture of how would be distributed if (using t with df = 3).

If the insert does not save power, the chance we would find such strong or stronger evidence than we got that it does is between 5% and 10%. This is assuming all conditions were met and the data were obtained in a proper fashion.

I1) HT for standard deviation use:

Hypothesis test for population variance: Get critical value(s) from table, and use where is from .

Picture of how would be distributed assuming . Note the tails are each .025. df = 3.

From A:

If the standard deviation is 50, the chance we would find such strong or stronger evidence than we got that it is not 50 is between 20% and 100%. This is assuming all conditions were met and the data were obtained in a proper fashion.

I2)

Picture of how would be distributed assuming . Note the tail is .05.

df = 3.

From A:

If the standard deviation is not over 50, the chance we would find such strong or stronger evidence than we got that it is over 50 is between 10% and 90%. This is assuming all conditions were met and the data were obtained in a proper fashion.

J) This is not a HT or a CI, it is a probability question, we need to find the area under the curve.

K) CI for proportion use:

1 sample proportion / z / sample proportion, p’=number of successes / n / population proportion, p, in Ho / HT: CI:

L) This is not a HT or a CI, it is a probability question, we need to find the area under the curve.

M1) Difference of two proportions, use:

Difference of proportions (percentages, or probabilities of success) from 2 samples / z / the difference of the two sample proportions subtracted in appropriate order / difference in population proportions in Ho (often 0) (0 and non 0 differences have different standard deviations see  / HT(0 case): where
CI & HT(non 0 case):

Picture of how all would be distributed if . The best evidence that Ha is true is in the shaded part that is in both tails. The total shaded areais.05.

If the percentages are the same, the chance we would find such strong or stronger evidence than we got they differ is .7188. This is assuming all conditions were met and the data were obtained in a proper fashion.

M2) or

Picture of how all would be distributed if . The best evidence that Ha is true is in the right tail of .05.

If the percentage is not greater for the other company, the chance we would find such strong or stronger evidence than we got that it is higher is .3594. This is assuming all conditions were met and the data were obtained in a proper fashion.

N) Matched pairs, use:

From earlier:

O)Difference of two proportions, use:

“other – original”

P) One mean, use:

1 sample mean / t, df = n-1 / sample mean / population mean in Ho /

From A)

945 945

Q1)HT for proportion use:

1 sample proportion / z / sample proportion, p’=number of successes / n / population proportion, p, in Ho / HT: CI:

Picture of how all p’ s would be distributed if . The best evidence for is in both tails. The total area of both tails is .05.

If the percentage is 1/3, the chance we would find such strong or stronger evidence than we got that it is not 1/3 is .5842. This is assuming all conditions were met and the data were obtained in a proper fashion.

Q2)

Picture of how all p’ s would be distributed if . The best evidence for is in the right tail of .05.

If the percentage is not over 1/3, the chance we would find such strong or stronger evidence than we got that it is over 1/3 is .2912. This is assuming all conditions were met and the data were obtained in a proper fashion.

R1) One mean, use:

1 sample mean / t, df = n-1 / sample mean / population mean in Ho /

Picture of how all ’s would be distributed if . The best evidence that is in both tails. Each tail is .025.

From earlier:

If the mean was 1000, the chance we would find such strong or stronger evidence than we got that it is not 1000 is between 10% and 20%. This is assuming all conditions were met and the data were obtained in a proper fashion.

R2)

Picture of how all ’s would be distributed if . The best evidence that is in the left tail of .05.

From earlier:

If the mean was at least 1000, the chance we would find such strong or stronger evidence than we got that it is less than 1000 is between 5% and 10%. This is assuming all conditions were met and the data were obtained in a proper fashion.

S) Difference of two means from independent samples use:

“Other – Original” or

T) z

4. Show two characteristics are related, use: O and E stuff : Right tails only. Use =

:two characteristics are independent : they are related

df=(r-1)(c-1)

E’s are found by (row total)(column total)/(grand total)

color and year independent

color and year related

O’s / black / white / blue / tan / Totals
2000 / 25 / 35 / 10 / 10 / 80
2001 / 60 / 60 / 16 / 24 / 160
2002 / 32 / 30 / 11 / 7 / 80
totals / 117 / 125 / 37 / 41 / 320
E’s / black / white / blue / tan
2000 / / / /
2001 / / / /
2002 / / / /

Picture of how all ’s would be distributed if color and year were independent. df =(4-1)(3-1) = 6

If there was no relationship, the chance we would find such strong or stronger evidence than we got that there is between 10% and 90%. This is assuming all conditions were met and the data were obtained in a proper fashion.

5. Showing data is not distributed a certain way, use: O and E stuff : Right tails only. Use =

: data distributed a certain way : its not distributed that way

df=number of categories – 1

E’s are found using np where p is the probability of being in a category in Ho

not as above

black / white / blue / tan / Total
O’s / 25 / 35 / 10 / 10 / 80
E’s / 80(.35) = 28 / 80(.35) = 28 / 80(.15) = 12 / 80(.15) = 12 / 80

Picture of how all ’s would be distributed if color and year were independent. df =(4-1) = 3

== NO

If the colors were distributed 35-35-15-15, the chance we would find such strong or stronger evidence than we got that the data is not distributed that way is between 10% and 90%. This is assuming all conditions were met and the data were obtained in a proper fashion.

6. A)

B) =.955

C):

YES

D)= =

or Plugging in x = 0 we get y = 19.2 and plugging in x = 14 we get y = 61.2. Next we plot the points and draw the line in the graph.

or 61.2

F) or 61.2