Question

A manufacturer must test that his bolts are 4.00cm long when they come off the assembly line. He must recalibrate his machines if the bolts are too long or too short. After sampling 81 randomly selected bolts off the assembly line, he calculates the sample mean to be 3.97cm. He knows that the population standard deviation is 0.14cm. Assuming a level of significance of 0.01, is there sufficient evidence to show that the manufacturer needs to recalibrate the machines

Answer 1

The p-value(0.0538) is greater than the given significance level of 0.01. So, there is no sufficient evidence to show that the manufacturer needs to recalibrate the machines.

What is the test statistics formula for finding the p-value approach of hypothesis testing?

The test statistics formula for finding the z-score is,

z = (X - μ)/(σ/√n)

Where,

X = sample mean

μ = population mean

σ = population standard deviation

n = sample size

With this z-score, the p-value is calculated. If p > α, then the null hypothesis is not rejected, and if p < α, then the null hypothesis is rejected.

Calculation:

It is given that,

The population mean μ = 4.00

The sample size n = 81

The sample mean X = 3.97

The standard deviation σ = 0.14

and the significance level α = 0.01

So, the hypothesis is:

The null hypothesis: H0: μ = 4

The alternative hypothesis: Ha: μ ≠ 4

Thus, the test statistic value is,

z = (X - μ)/(σ/√n)

  = (3.97 - 4)/(0.14/√81)

  = - 0.03/0.015

  = -1.928

So, for z = -1.928, the p-value is 0.053855

Since p > 0.01, there is no sufficient evidence to show that the manufacturer needs to recalibrate the machines.

Learn more about hypothesis testing here:

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