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Anton [14]
2 years ago
12

A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of seco

ndary school teachers in a large school district. The mean of the salaries of a random sample of 26 elementary school teachers is $48,250 and the sample standard deviation is $3900. The mean of the salaries of 24 randomly selected secondary school teachers is $45,630 with a sample standard deviation of $5530. At ???? = 0.05, can it be concluded that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers?
Mathematics
1 answer:
Brut [27]2 years ago
5 0

Answer:

t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921  

df=n_{A}+n_{B}-2=26+24-2=48

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

p_v =P(t_{(48)}>1.921)=0.0303

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

Step-by-step explanation:

Data given and notation

\bar X_{A}=48250 represent the mean of elementary teachers

\bar X_{B}=45630 represent the mean for secondary teachers

s_{A}=3900 represent the sample standard deviation for elementary teacher

s_{B}=5530 represent the sample standard deviation for secondary teachers

n_{A}=26 sample size selected

n_{B}=24 sample size selected  

\alpha=0.05 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 of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers, the system of hypothesis would be:

Null hypothesis:\mu_{A}-\mu_{B}\leq 0

Alternative hypothesis:\mu_{A}-\mu_{B}>0

We don't know the population deviations, so for this case is better apply a t test to compare means, and the statistic is given by:

t=\frac{(\bar X_{A}-\bar X_{B})}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}} (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

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

t=\frac{(48250-45630)}{\sqrt{\frac{3900^2}{26}+\frac{5530^2}{24}}}}=1.921  

P-value

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

df=n_{A}+n_{B}-2=26+24-2=48

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

p_v =P(t_{(48)}>1.921)=0.0303

Conclusion

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the mean for elementary school teachers is significantly higher than the mean for secondary teachers at 5% of significance

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