What is the square root of 100?

A research firm tests the miles-per-gallon characteristics of three brands of gasoline. Because of different gasoline performanc

Anni [7]

Answer:

A.) At α = 0.05, is there a significant difference in the mean miles-per-gallon characteristics of the three brands of gasoline.

B.)The advantage of attempting to remove the block effect is

The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.

Step-by-step explanation:

Using Two-way ANOVA method

Given problem

Observation I II III Row total (xr)

A 18 21 20 59

B 24 26 27 77

C 30 29 34 93

D 22 25 24 71

E 20 23 24 63

Col total (xc) 114 124 129 367

∑x²=9233→(A)

∑x²c/r

=1/5(114²+124²+129²)

=1/5(12996+15376+16641)

=1/5(45013)

=9002.6→(B)

∑x²r/c

=1/3(59²+77²+93²+71²+67²)

=1/3(3481+5929+8649+5041+4489)

=1/3(27589)

=9196.3333→(C)

(∑x)²/n

=(367)²/15

=134689/15

=8979.2667→(D)

Sum of squares total

SST=∑x²-(∑x)²/n

=(A)-(D)

=9233-8979.2667

=253.7333

Sum of squares between rows

SSR=∑x²r/c-(∑x)²/n

=(C)-(D)

=9196.3333-8979.2667

=217.0667

Sum of squares between columns

SSC=∑x²c/r-(∑x)²/n

=(B)-(D)

=9002.6-8979.2667

=23.3333

Sum of squares Error (residual)

SSE=SST-SSR-SSC

=253.7333-217.0667-23.3333

=13.3333

ANOVA table

Source Sums Degrees Mean Squares

of Variation of Squares of freedom

SS DF MS F p-value

B/ w SSR=217.0667 4 MSR=54.2667 32.56 0.0001

rows

B/w SSC=23.3333 c-1=2 MSC=11.6667 7 0.01

columns

Error (residual)SSE=13.3333 (r-1)(c-1)=8 MSE=1.6667

Total SST=253.7333 rc-1=14

Conclusion:

1. F for between Rows

The critical region for F(4,8) at 0.05 level of significance=3.8379

The calculated F for Rows=32.56>3.8379

Therefore H0 is rejected

2. F for between Columns

The critical region for F(2,8) at 0.05 level of significance=4.459

We see that the calculated F for Colums=7>4.459

therefore H0 is rejected,and concluded that there is significant differentiating between columns

Part B:

To analyze the data for completely randomized designs click on anova two factor without replication in the data analysis dialog box of the excel spreadsheet.

The following table is obtained

Source DF Sum Mean F Statistic

(df1,df2) of Square (SS) Square (MS) P-value

Factor A 1 1496.5444 1496.5444 769.6514 0.001297

Rows

Factor B - 2 19.4444 9.7222 5 0.1667

Columns

Interaction

AB 2 3.8889 1.9444 0.1013 0.9045

Error 12 230.4 19.2

Total 17 1750.2778 102.9575

Factor - A- Rows

Since p-value < α, H0 is rejected.

Factor - B- Columns

Since p-value > α, H0 can not be rejected.

The averages of all groups assume to be equal.

Interaction AB

Since p-value > α, H0 can not be rejected.

The advantage of attempting to remove the block effect is

The completely randomized designs does not prove that H0 is incorrect only that it cannot be rejected.

3 0

3 years ago

5.) what is the number symbol of four is to eight

zalisa [80]

Answer:

4:8

Step-by-step explanation:

I think

7 0

3 years ago

Babies born after 40 weeks gestation have a mean length of 52.2 centimeters (about 20.6 inches). Babies born one month early hav

Sveta_85 [38]

Answer:

a) $Z = -2.88$

b) $Z = -0.96$

c) 40 weeks gestation babies

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

a. Find the standardized score (z-score), relative to all U.S. births, for a baby with a birth length of 45 cm.

Here, we use $\mu = 52.2, \sigma = 2.5$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{45 - 52.2}{2.5}$

$Z = -2.88$

b. Find the standardized score of a birth length of 45 cm. for babies born one month early, using 47.4 as the mean.

Here, we use $\mu = 47.4, \sigma = 2.5$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{45 - 47.4}{2.5}$

$Z = -0.96$

c. For which group is a birth length of 45 cm more common?

For each group, the probability is 1 subtracted by the pvalue of Z.

Z = -2.88 has a lower pvalue than Z = -0.96, so for Z = -2.88 the probability 1 - pvalue of Z will be greater. This means that for 40 weeks gestation babies a birth length of 45 cm is more common.

3 0

3 years ago

Is 5x-3>5x-3 an identity or a contradiction

sesenic [268]

Answer:

contradiction

Step-by-step explanation:

they are both the same thing so how can they contradict? they can because the sign in the middle is greater than. if you put 1 in as x, you will get 2 is greater than 2 which does not make sense

6 0

3 years ago

Read 2 more answers

Help anyone I have like 15 more minuets to finish

Naya [18.7K]

The answer is 45 it's the easiest

5 0

3 years ago