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julia-pushkina [17]
3 years ago
15

A group of 300 housewives was interviewed to determine if there is a preference for one of two detergents. Detergent A was favor

ed by 135 housewives; the others favored Detergent B. Which procedure would you perform to ascertain if the data provide sufficient evidence to indicate a difference in preference for the two detergents
Mathematics
1 answer:
san4es73 [151]3 years ago
3 0

Answer: Using Chi-square.

Step-by-step explanation: To check if the provided data show any differences in preference for the two detergents (i.e., two variables), a Chi-square can be deployed, as it is used to test the goodness of fit, and test if two variables are related.

If the value obtained is less than 0.05, it means that there is statistical difference between the preference for the two detergents.

However, if the obtained value is higher than 0.05, it translates that there is no statistical difference between the preference for the two detergents.

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I must make some assumptions here about what you may have meant by your "<span>linear equation y=3x−5y=3x−5 y equals 3 x , minus 5."

You've written "y=3x-5" three times on the same line of type.  Why is that?  

Let's change what you've typed to the following:

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A safe has a 4-digit lock code that does not include zero as a digit is repeated. What is the probability of a lock without all
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Answer:

t=\frac{120-100}{\frac{20}{\sqrt{10}}}=3.16  

p_v =2*P(t_{9}>3.16)=0.012  

If we compare the p value and a significance level assumed \alpha=0.05 we see that p_v so we can conclude that we can reject the null hypothesis, and the true mean is significantly different from 100 at 5% of significance.  

Step-by-step explanation:

Data given and notation

\bar X=120 represent the sample mean  

s=20 represent the standard deviation for the sample

n=10 sample size  

\mu_o =100 represent the value that we want to test  

\alpha 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 to be tested  

We need to conduct a hypothesis in order to determine if the mean is different from 100, the system of hypothesis would be:  

Null hypothesis:\mu = 100  

Alternative hypothesis:\mu \neq 100  

Compute the test statistic  

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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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".  

We can replace in formula (1) the info given like this:  

t=\frac{120-100}{\frac{20}{\sqrt{10}}}=3.16  

Now we need to find the degrees of freedom for the t distirbution given by:

df=n-1=10-1=9

Conclusion

Since is a tao tailed test the p value would be:  

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If we compare the p value and a significance level assumed \alpha=0.05 we see that p_v so we can conclude that we can reject the null hypothesis, and the true mean is significantly different from 100 at 5% of significance.  

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