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# An experimental surgical procedure is being studied as an alternative to the old method. Both methods...

An experimental surgical procedure is being studied as an alternative to the old method. Both methods are considered safe. Five surgeons perform the operation on two patients matched by age, sex, and other relevant factors, with the results shown. The time to complete the surgery (in minutes) is recorded.

 Surgeon 1 Surgeon 2 Surgeon 3 Surgeon 4 Surgeon 5 Old way 40 60 31 38 64 New way 32 44 35 36 53

(a-1) Calculate the difference between the new and the old ways for the data given below. Use α = .025. (Negative values should be indicated by a minus sign.)

 X1 X2 X1 - X2 Surgeon Old Way New Way Difference 1 40 32 2 60 44 3 31 35 4 38 36 5 64 53

(a-2) Calculate the mean and standard deviation for the difference. (Round your mean answer to 1 decimal place and standard deviation answer to 4 decimal places.)

 Mean Standard Deviation

(a-3)
Choose the right option for H0: μd ≤ 0; H1: μd> 0.

• Reject if tcalc > 2.776445105

• Reject if tcalc < 2.776445105

(a-4) Calculate the value of tcalc. (Round your answer to 4 decimal places.)

t
calc

(b-1) Is the decision close? (Round your answer to 4 decimal places.)

The decision is  (Click to select)  close  not close  .

The p-value is  .

(b-2) The new way is better than the old.

• No

• Yes

(b-3) The difference is significant.

• No

• Yes

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(a - 1)

 X1 Old way X2 New way X1 - X2 40 32 8 60 44 16 31 35 -4 38 36 2 64 53 1

(a - 2)

From the values of X1 - X2, the following statistics are calculated:

(i)

Mean = = 4.6

(ii)

Standard Deviation = sd = 7.6681

(a - 3)

Correct option:

H0: μd ≤ 0; H1: μd > 0.

=0.025

df = 5 - 1 = 4

By Technology, critical value of t = 2.776445105

So,

Decision Rule:

• Reject if tcalc > 2.776445105

(a - 4)

Test Statistic is got as follows:

(b -1)

Correct option:

not close

By Technology,

The p value is 0.1254

(b - 2)

Correct option:

No

(b - 3)

Correct option:

No

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