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?

If we compare the p value and the significance level assumed we see that 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  

represent the sample mean

represent the sample standard deviation  

sample size  

represent the value that we want to test  

represent the significance level for the hypothesis test.  

z 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 1.5 inch, the system of hypothesis would be:  

Null hypothesis:  

Alternative hypothesis:  

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:  

(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:  

P-value  

The degrees of freedom are given by:

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

Conclusion  

If we compare the p value and the significance level assumed we see that 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.  

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.

m a t h w a y. c o m

step-by-step explanation:

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation: