Mathematics, 14.04.2020 16:37 mirandatorres2601

You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 30 night students, and the sample mean GPA is 2.34 with a standard deviation of 0.456. You sample 25 day students, and the sample mean GPA is 2.59 with a standard deviation of 0.482. Test the claim using a 10% level of significance. Assume the sample standard deviations are unequal and that GPAs are normally distributed.

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Answer from: ste78

$t=\frac{2.59-2.34}{\sqrt{\frac{0.482^2}{25}+\frac{0.456^2}{30}}}}=1.963$

$df=n_{A}+n_{B}-2=30+25-2=53$

$p_v =2*P(t_{(53)}1.963)=0.0549$

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, and we can conclude that the two means are different at 10% of significance.

Step-by-step explanation:

Data given and notation

$\bar X_{A}=2.34$ represent the mean for the sample of night students

$\bar X_{B}=2.59$ represent the mean for the sample of day students

$s_{A}=0.456$ represent the sample standard deviation for the sample of night students

$s_{B}=0.482$ represent the sample standard deviation for the sample of day students

$n_{A}=30$ sample size selected for the night students

$n_{B}=25$ sample size selected for the day students

$\alpha=0.1$ 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 two means are equal or no, the system of hypothesis would be:

Null hypothesis: $\mu_{A} = \mu_{B}$

Alternative hypothesis: $\mu_{A} \neq \mu_{B}$

If we analyze the size for the samples both are <= than 30 and we don't know the population deviations, for this case is better apply a t test to compare means, and the statistic is given by:

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

$t=\frac{2.59-2.34}{\sqrt{\frac{0.482^2}{25}+\frac{0.456^2}{30}}}}=1.963$

P-value

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

$df=n_{A}+n_{B}-2=30+25-2=53$

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

$p_v =2*P(t_{(53)}1.963)=0.0549$

Conclusion

Answer from: Quest

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step-by-step explanation:

Answer from: Quest

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step-by-step explanation:

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