Mathematics: At a large university, a sample of 52 evening students was selected in order to determine whether the average age of the evening students is significantly different from 21. The average age of the students in the sample was 21.48 years. The sample standard deviation was 1.3 years. Determine whether or not the average age of the evening students is significantly different from 21. Use a 0.01 level of significance. Choose the t-statistic and the appropriate conclusion. a. t= 2.6626 ; reject the null

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At a large university, a sample of 52 evening students

Ответ:

wendelkristen

26.01.2022

Mathematics

$t=\frac{21.48-21}{\frac{1.3}{\sqrt{52}}}=2.6626$

$p_v =2*P(t_{51}2.6626)=0.0103$

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis

And the answer would be

d. t= 2.6626; fail to reject the null hypothesis

Step-by-step explanation:

Data given

$\bar X=21.48$ represent the sample mean

s=1.3 represent the sample standard deviation

n=52 sample size

$\mu_o =21$ represent the value that we want to test

$\alpha=0.01$ 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 true mean for the the age is different from 21, the system of hypothesis would be:

Null hypothesis: $\mu =21$

Alternative hypothesis: $\mu \neq 21$

Since we don't know the population deviation, 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)

Calculate the statistic

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

$t=\frac{21.48-21}{\frac{1.3}{\sqrt{52}}}=2.6626$

P-value

The degrees of freedom are given by:

df = n-1=52-1=51

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

$p_v =2*P(t_{51}2.6626)=0.0103$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v\alpha$ so we can conclude that we have enough evidence to FAIL reject the null hypothesis

And the answer would be:

d. t= 2.6626; fail to reject the null hypothesis

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