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attashe74 [19]
3 years ago
7

A researcher was interested in comparing the salaries of female and male employees of a particular company. independent random s

amples of female employees (sample 1) and male employees (sample 2) were taken to calculate the mean salary, in dollars per week, for each group. A 95% confidence interval for the difference, µ1- µ2, between the mean weekly salary of all female employees and the mean weekly salary of all male employees was determined to be (-$110,-$10). Interpret the results.
Mathematics
1 answer:
Aneli [31]3 years ago
5 0

Answer:

Based on  given data, we can say with 95% confidence that the female employees at this company average between $110 less and $10 less per week than the male employees.

Step-by-step explanation:

Sample 1 for female

Sample 2 for male

Confidence = 90%

Difference in mean weekly salary of all female employees and the mean weekly salary of all male employees  = µ₁- µ₂

So based on above given data, we can say with 95% confidence that the female employees at this company average between $110 less and $10 less per week than the male employees.

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The mean annual premium for automobile insurance in the United States is $1503 (Insure website, March 6, 2014). Being from Penns
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Answer:

If we compare the p value and the significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean annual premium in Pennsylvania is significantly lower than the national mean annual premium of 1503 at 5% of significance.

Step-by-step explanation:

1) Data given and notation  

\bar X=1440 represent the mean annual premium value for the sample  

s=165 represent the sample standard deviation for the sample  

n=25 sample size  

\mu_o =1503 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.  

We need to conduct a hypothesis in order to check if the mean annual premium in Pennsylvania is lower than the national mean annual premium, the system of hypothesis would be:  

Null hypothesis:\mu \geq 1503  

Alternative hypothesis:\mu < 1503  

If we analyze the size for the sample is < 30 and 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{1440-1503}{\frac{165}{\sqrt{25}}}=-1.909    

P-value

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

df=n-1=25-1=24  

Since is a one side left tailed test the p value would be:  

p_v =P(t_{(24)}  

Conclusion  

If we compare the p value and the significance level for example \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean annual premium in Pennsylvania is significantly lower than the national mean annual premium of 1503 at 5% of significance.  

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