A manufacturing company produces valves in various sizes and shapes. One particular valve plate is supposed to have a tensile strength of 5 pounds per millimeter (lbs/mm). The company tests a random sample of 42 such valve plates from a lot of 650 valve plates. The sample mean is a tensile strength of 5.0611 lbs/mm, and the population standard deviation is 0.2803 lbs/mm. Use α = .10 and test to determine whether the lot of valve plates has an average tensile strength of 5 lbs/mm.

If we compare the p value and the significance level given we see that so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

Step-by-step explanation:

Data given and notation  

represent the sample mean

represent the population standard deviation for the sample  

sample size  

represent the value that we want to test

represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

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 true mean is different from 5, the system of hypothesis would be:  

Null hypothesis:  

Alternative hypothesis:  

If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

 (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

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

P-value

Since is a two sided test the p value would be:  

Conclusion  

If we compare the p value and the significance level given we see that so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

i think that you should ask google for the answer to that problem because if we don't know it google does

option a is correct.

step-by-step explanation:

scale factor: it is used to find proportional measurements.

proportions means having the same ratio.

scale factor is the ratio of the model measurement to the actual measurement in a simplest form.

or we can say that the ratio of any two corresponding lengths in two similar geometric figures

if the scale factor is more than 1, then it is an enlargement,

if the scale factor less than 1, then it is a reduction.

given the two geometric figure:

by definition of proportions, we have