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rjkz [21]
2 years ago
13

A home building company routinely orders standard interior doors with a height of 80 inches. Recently the installers have compla

ined that the doors are not the standard height. The quality control inspector for the home building company is concerned that the manufacturer is supplying doors that are not 80 inches in height. In an effort to test this, the inspector is going to gather a sample of the recently received doors and measure the height. The alternative hypothesis for the statistical test to determine if the doors are not 80 inches is
Mathematics
1 answer:
Helga [31]2 years ago
3 0

Answer: Alternative Hypothesis

Ha : u ≠ 80 inches ( mean not equal to 80 inches)

Step-by-step explanation:

The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean.

While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.

For the case above, the mean door height is 80 inches

H0 : u = 80 inches

Ha : u ≠ 80 inches ( mean not equal to 80 inches)

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Answer:

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

p_v =2*P(t_{(1699)}>1.971)=0.049  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation  

\bar X=669 represent the sample mean

s=732 represent the sample standard deviation

n=1700 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0. represent the significance level for the hypothesis test.  

t 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 mean is different from 634, the system of hypothesis would be:  

Null hypothesis:\mu = 634  

Alternative hypothesis:\mu \neq 634  

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=1700-1=1699  

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

p_v =2*P(t_{(1699)}>1.971)=0.049  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

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we have

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Rewrite as perfect squares

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