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#6 please

17%ZUWritten%20HW5.pdf ::: AppsFarulty Directory | M D New TabCaesepcia Ruld Guir ieCirVE. IIE IIIasses aIE SIIOWI below in milligrams. Determine the 95% confidence interval for the difference in the mean masses of the tissues between the two group Group 1: 60 62 63 51 54 61 69 60 67 57 Group 2: 58 66 60 66 60 68 61 6. A shoe engineer wants to estimate the comfort level for a new material used for soles. The testing approach examines the amount of total time that the shoes are worn by 7 people till the point where they complain of discomfort. The following amounts of time were recorded (in months) Find a 95% confidence interval for the mean amount of time If you had the same variance and mean from sample of 50 people, how would your interval change? a. b. 10.4 9.8 10.0 10.2 9.8 9.6 10.2 7. A hardness testing machine presses a rod with a pointed tip into a metal specimen with a known force. By measuring the depth of the depression caused by the tip, the hardness of the specimen is determined. Two different tips are available for this machine. Although the precision of the measurements produced by the two tips is the same, it is suspected that one tip produces different mean hardness ratings than the other Type here to search 1011/2017

Explanation / Answer

a) I am using R software to do some basic calculations.

First we can create a vector holding the values given in the table as below:
Data <- c(9.8,10.4,10.0,9.6,10.2,9.8,10.2)
We can calculate the mean and standard deviation using the mean() and sd() functions respectively as shown below.

Sample mean = mean(Data) = 10
Sample sd = sd(Data) = 0.2828427

We can calculate the Standard error as below:

Standard error = sd/sqrt(n) = 0.2828427/sqrt(7) = 0.1069045
Degrees of freedom = n - 1 = 7 - 1 = 6

Confidence interval is given as 95%
So significance level = 0.05
So critical probability = 1 - 0.05/2 = 0.975
Degrees of freedom = 6

Now Critical value of the test statistic can be found out using the t table, which is 2.447

Now we need to find the margin of error as below:
Margin of error = critical value * standard error = 2.447 * 0.1069045 = 0.2615953

Confidence interval = sample statistic +- margin of error
= 10 +- 0.2615953
= (9.738405,10.2616)

b) If we have the same variance and mean from a sample of 50 people, the values for standard error and critical value will change as below:
Standard error = sd/sqrt(n) = 0.2828427/sqrt(50) = 0.04
Degrees of freedom = n - 1 = 50 - 1 = 49

Now Critical value of the test statistic can be found out using the t table, which is 2.01

Margin of error = critical value * standard error = 2.01 * 0.04 = 0.0804

Confidence interval = sample statistic +- margin of error
= 10 +- 0.0804
= (9.9196,10.0804)

So confidence interval has become narrower.

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