Mathematics, 04.12.2019 04:31 karsynraine8224

Amanufacturer claims that the average ten- sile strength of thread a exceeds the average tensile strength of thread b by at least 12 kilograms. to test this claim, 50 pieces of each type of thread were tested under similar conditions. type a thread had an aver- age tensile strength of 86.7 kilograms with a standard deviation of 6.28 kilograms, while type b thread had an average tensile strength of 77.8 kilograms with a standard deviation of 5.61 kilograms. test the manufacturer’s claim using a 0.05 level of significance

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Answer from: Nay25933

$t=\frac{(86.7-77.8)-12}{\sqrt{\frac{6.28^2}{50}+\frac{5.61^2}{50}}}}=-2.603$

$p_v =P(t_{(98)}-2.603)=0.995$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and the difference between the average ten- sile strength of thread A and the average ten- sile strength of thread B is not signficantly higher than 12 at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X_{A}=86.7$ represent the mean for the sample A

$\bar X_{B}=77.8$ represent the mean for the sample B

$s_{A}=6.28$ represent the sample standard deviation for the sample A

$s_{B}=5.61$ represent the sample standard deviation for the sample B

$n_{A}=50$ sample size selected A

$n_{B}=50$ sample size selected B

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

p_v represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the average ten- sile strength of thread A exceeds the average tensile strength of thread B by at least 12 kilograms., the system of hypothesis would be:

Null hypothesis: $\mu_{A}-\mu_{B}\leq 12$

Alternative hypothesis: $\mu_{A}-\mu_{B}12$

We don't know the population deviations, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{(\bar X_{A}-\bar X_{B})-12}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{(86.7-77.8)-12}{\sqrt{\frac{6.28^2}{50}+\frac{5.61^2}{50}}}}=-2.603$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{A}+n_{B}-2=50+50-2=98$

Since is a one side test the p value would be:

$p_v =P(t_{(98)}-2.603)=0.995$

Conclusion

Answer from: Quest

1.17+3.99a. you combine the terms with a to create 3.99 and the 1.17 stays because it has no like terms or a variable

Answer from: Quest

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