Mathematics, 22.04.2020 02:11, 4804113260

Bob is the owner of a home improvement store. He has hired you to check his machine’s calibration prior to starting production on a large order. To check this, you set the machine to create 1.5 inch bolts and manufacture a random sample of 200 bolts. That sample of bolts has an average length of 1.521 inches with a standard deviation of 0.204 inches. Does this sample provide convincing evidence that the machine is working properly or should it be shut down for repairs?

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answered: liqliq223

$t=\frac{1.521-1.5}{\frac{0.204}{\sqrt{200}}}=1.456$

p_v =2*P(z1.456)=0.145

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the true mean is not different from the specification at 1% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=1.521$ represent the sample mean

s=0.204 represent the sample standard deviation

n=200 sample size

$\mu_o =1.5$ represent the value that we want to test

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

z 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 true mean is different from 1.5 inch, the system of hypothesis would be:

Null hypothesis: $\mu =1.5$

Alternative hypothesis: $\mu \neq 1.5$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-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:

$t=\frac{1.521-1.5}{\frac{0.204}{\sqrt{200}}}=1.456$

P-value

The degrees of freedom are given by:

df = n-1= 200-1=199

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

p_v =2*P(z1.456)=0.145

Conclusion

answered: Juanga122

Parameter:

Null hypothesis: μ = 1.5 (the machines work as needed)

Alternative hypothesis: μ ≠ 1.5 (The machines don't work properly)

Since we don't know the population deviation, we will apply a t-test to compare the actual mean to the reference value

Conditions:

Simple random sample: The problem states the sample was chosen at random.

Independence: You can assume there are more than 10(200) = 2000 screws.

Normality: (200 ≥ 30) the sample is large enough for sampling distribution to assume Normality

Calculations:

Since the conditions are met we will carry out a T-test using a calculator for μ≠μ0

μ = population mean = 1.5

σ= standard dev = 0.204

xbar = sampe mean = 1.521

n = sample size= 200

After adding all of this data into the calculator in the T-test program we get a p-value of 0.147

Conclusion:

We will assume a 0.05 sig level for our conclusion.

Since 0.147 > 0.05 we will fail to reject the null hypothesis meaning that we have enough evidence to show that the machines work as needed.

answered: Guest

histogram or dot plot?

step-by-step explanation:

Iwont to know i usually have more teeth than reptiles. which type of graph would allow us to compare

answered: Guest

c is the answer

step-by-step explanation:

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Bob is the owner of a home improvement store. He has hired you to check his machine’s calibration pr...