In a mechanical testing lab, Plexiglass strips are stretched to failure. Assume

Question

In a mechanical testing lab, Plexiglass strips are stretched to failure. Assume that the change in length in mm before breaking has a normal distribution. A random sample of 6 observations of the length change yielded the following data. 5.1 5.3 6.2 6.4 5.6 5.8 Dose the data substantiate the claim that the mean change in length is greater than 5.7 mm? Answer using the p-value. Does the data substantiate the claim that the mean change in length is less than 5.7 mm? Answer using the p-value. Does the data substantiate the claim that the mean change in length is not 5.7 mm? Answer using the p-value. Answer parts 1, 2, 3 if the standard deviation of the length change is known to be 0.45 mm and use a 5% significance level instead of the p-value.

Explanation / Answer

We use excel functions to get the mean and standard deviation values:

=average(data set)

=stdev(data set)

x

5.1

5.3

6.2

6.4

5.6

5.8

x bar =

5.7333

s =

0.5046

Question 1)

H0: µ = 5.7

H1: µ > 5.7

Level of significance = 0.05

Test Statsitics:

z = ( x bar – Mew)/(s/sqrt(n))

= (5.7 – 5.7333)/(0.45/sqrt(6))

= -0.18

p-value =tdist(0.162,5,1)

p-value = 0.4388

Here the p-value is greater than the level of significance (0.4388 > 0.05); we fail to reject the null hypothesis.

There is not sufficient evidence to conclude that the length is greater than 5.7

  

Question 2)

H0: µ = 5.7

H1: µ < 5.7

Level of significance = 0.05

Test Statsitics:

t = ( x bar – Mew)/(s/sqrt(n))

= (5.7 – 5.7333)/(0.5046/sqrt(6))

= -0.162

Df = n - 1 = 6 – 1 = 5

p-value =tdist(0.162,5,1)

p-value = 0.4388

Here the p-value is greater than the level of significance (0.4388 > 0.05); we fail to reject the null hypothesis.

There is not sufficient evidence to conclude that the length is less than 5.7

Question 3)

H0: µ = 5.7

H1: µ not equal 5.7

Level of significance = 0.05

Test Statsitics:

t = ( x bar – Mew)/(s/sqrt(n))

= (5.7 – 5.7333)/(0.5046/sqrt(6))

= -0.162

Df = n - 1 = 6 – 1 = 5

p-value =tdist(0.162,5,2)

p-value = 0.8766

Here the p-value is greater than the level of significance (0.8766 > 0.05); we fail to reject the null hypothesis.

There is not sufficient evidence to conclude that the length is not equal to 5.7

Question 4) Part I

H0: µ = 5.7

H1: µ > 5.7

Level of significance = 0.05

Test Statsitics:

z = ( x bar – Mew)/(sigma/sqrt(n))

= (5.7 – 5.7333)/(0.45/sqrt(6))

= -0.18

p-value =normsdist(-0.18)

p-value = 0.4286

Here the p-value is greater than the level of significance (0.4286 > 0.05); we fail to reject the null hypothesis.

There is not sufficient evidence to conclude that the length is greater than 5.7

Question 4) Part II

H0: µ = 5.7

H1: µ < 5.7

Level of significance = 0.05

Test Statsitics:

z = ( x bar – Mew)/(sigma/sqrt(n))

= (5.7 – 5.7333)/(0.45/sqrt(6))

=-0.18

p-value =1-normsdist(-0.18)

p-value = 0.5714

Here the p-value is greater than the level of significance (0.5714 > 0.05); we fail to reject the null hypothesis.

There is not sufficient evidence to conclude that the length is less than 5.7

Question 4) Part III

H0: µ = 5.7

H1: µ not equal 5.7

Level of significance = 0.05

Test Statsitics:

z = ( x bar – Mew)/(sigma/sqrt(n))

= (5.7 – 5.7333)/(0.45/sqrt(6))

= -0.18

p-value =2*normsdist(-0.18)

p-value = 0.8572

Here the p-value is greater than the level of significance (0.8572 > 0.05); we fail to reject the null hypothesis.

There is not sufficient evidence to conclude that the length is not equal to 5.7

x

5.1

5.3

6.2

6.4

5.6

5.8

x bar =

5.7333

s =

0.5046

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