Question

1. You work for a company that manufactures steel beams, and you wish to make sure the beams are up to industry standards. The average steel beam needs to have a tensile strength of 400 MPa (mega pascals). The company hopes that the beams are stronger than average, because they are for a very important client. You have been tasked with checking that the beams are stronger than average. You have tested the tensile strength of the beams you manufacture, and have recorded the following results. X 463 MPa s 123 n 15 A) State your null and alternative hypotheses B) Carry out the test with a-0.05 C) Are the beams significantly stronger than average at the 99% confidence level? D) Would you suggest these beams be sold to your important client? Explain your answer. Report the p-value and state the conclusion of the hypothesis tests, whose results are given belovw 2. A) A two-tailed test with a 0.05, and z-sta-1.94 B) A right-tailed test with a -0.10, and z-stat 1.29 C) A left-tailed test with α-.01, and z-stat -2.75

Answer 1

Given Information,

n= 15, x=463. s= 123

A)

Our hypothesis is,

  

i.e Beams are stronger than average (Rigth tailed test)

H1 : μ > 400

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B) Since, sample size is small (n<30) and population standard deviation () is unknown, so t-test is use here.

-n 123-100 = 31,63846 t = 1.983723 /Vn 23/V15

Test statistics is 1.9837

Degree of freedom = n-1 = 15-1 = 14

Critical t value at 0.05 level of significance for 14 degree of freedom is 1.761 (Obtain from t-table)

Since,

Conclusion:  Since, null hypothesis is rejected. Therefore, we conclude that there is enough evidence to support the claim that Beams are stronger than average.

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C) Confidence level = 99%

Critical t value at 0.01 level of significance for 14 degree of freedom is 2.624 (From t-table)

Since,

tcalculated < 2.624. Ho is not rejected

Conclusion:  Since, null hypothesis is not rejected at 99% confidence level. Therefore, we conclude that there is not enough evidence to support the claim that Beams are stronger than average.

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2.

A). Given,

= 1.95, a 0.05

For two tailed test, p value is calulated in R as:

2*(1-pnorm(1.94)) =  0.0524

Since, p value is greater than the specified level of significance, p> 0.05, null hypothesis is not rejected.

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B)

Given,

For right tailed test, p value is calulated in R as:

(1-pnorm(1.29)) =   0.0985

Since, p value is less than the specified level of significance, p< 0.10, null hypothesis is rejected.

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C)

Given,

For left tailed test, p value is calulated in R as:

(1-pnorm(2.75)) =   0.00297

Since, p value is less than the specified level of significance, p< 0.01, null hypothesis is rejected.

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