statistics Question 2 A manufacturer is reconditioning old electronic devices. T

Question

statistics Question 2 A manufacturer is reconditioning old electronic devices. The manufacturer needs an component with an average resistance of 350 The manufacturer randomly collected the following data from his main provider named Providerl. Provider 1: 344 353 344 344 345 342 334 357 355 350 356 347 34 355 350 351 350 366 1- Give point estimates of mean and variance. 2-Can we consider, at = 5%, that the sample meets the Standard. 3- Discuss the assumptions, and verify their validity 4-What will be your answer if -1%. 5- F or the same electronic component, the manufacturer has now a second provider, named in what follows Provider 2. The manufacturer wants to test if the average resistance of the two providers are equal. The manufacturer randomly collected the following sample from Provider 2: 36 339 341 330 323 360 357 344 358 345 328 351 346 342 344 355 335 356 332 Compare the average resistance of the two samples at = 1%. What will be the manufacturer decision? a. b. 6- The manufacturer considers the components as nondefective if its resistance is between d 365 At = 1% compare the proportions of defective components provided by 335 an both providers. 7- Suppose that you are the manufacturer and based on the previous questions, discuss the quality of the components delivered to you and what will be your decision.

Explanation / Answer

point estiamtes are sample mean and sample variance.

Sample mean=sum/total

=344+353+344+344+345+342+334+357+355+350+356+347+341+355+350+351+350+366/18

=349.111

sample variance

=54.69281

sample sd=sqrt(54.69281)= 7.395459

Solution2:

Null hypothesis:

H0: =350

Alternative Hypothesis:

Ha: 350

alpha=5%=0.05

test statistic:

t=sample mean-population mean/sample sd/sqrt(n)

=349.111-350/ 7.395459/sqrt(18)

t =-0.50994

p-value = 0.6167

p>0.05

Fail to reject Null hypothesis

Accept Null hypothesis.

Conclusion:

there is suffcinet evidence at 5% level of significance to conclude that

sample meets standard.

Rcode for the same output was:

resistance <- c(344,353,344,344,345,342,334,357,355,350,356,347,341,355,350,351,350,366)

t.test(resistance,mu=350)

output:

One Sample t-test

data: resistance

t = -0.50994, df = 17, p-value = 0.6167

alternative hypothesis: true mean is not equal to 350

95 percent confidence interval:

345.4334 352.7888

sample estimates:

mean of x

349.1111

Solutionc:

Conduct shapiro test in R

Code:

resistance <- c(344,353,344,344,345,342,334,357,355,350,356,347,341,355,350,351,350,366)

shapiro.test(resistance)

Output:

Shapiro-Wilk normality test

data: resistance

W = 0.97545, p-value = 0.8916

intrepretation:

p-value = 0.8916

p>0.05

Data comes from normal distribution.

mean =median

Data comes from normal distribution.

T test assumptions are met

sample is simple random sample.

normality assumption is met.

Can conduct t.test.

Solution4:

alpha=1%=0.01

conduct t test in R

t.test(resistance,mu=350,conf.level=0.99)

Output:

One Sample t-test

data: resistance

t = -0.50994, df = 17, p-value = 0.6167

alternative hypothesis: true mean is not equal to 350

99 percent confidence interval:

344.0591 354.1631

sample estimates:

mean of x

One Sample t-test

data: resistance

t = -0.50994, df = 17, p-value = 0.6167

alternative hypothesis: true mean is not equal to 350

99 percent confidence interval:

344.0591 354.1631

sample estimates:

mean of x

349.1111

p=0.6167

p>0.01

Reject null hypothesis.

Accept alternative hypothesis.

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