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The security department of a factory wants to know whether the true average time required by the night guard to walk his round is 30 minutes. If, in a random sample of 32 rounds, the night guard averaged 30.8 minutes with a standard deviation of 1.5 minutes, determine whether this is sufficient evidence to reject the null hypothesis µ = 30 minutes in favor of the alternative hypothesis µ 6= 30 minutes, at the 0.01 level of significance. Conduct the test using the p-value approach. Provide detailed solutions in the four steps to hypothesis testing and state your conclusion in the context of the problem.

User Rutsky
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Answer:


t=(30.8-30)/((1.5)/(√(32)))=3.017


p_v =2*P(t_((31))>3.017)=0.0051

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 reject the null hypothesis, so we can conclude that the treu mean is different from 30 minutes at 1% of significance

Explanation:

Data given and notation


\bar X=30.8 represent the sample mean


s=1.5 represent the sample standard deviation


n=32 sample size


\mu_o =30 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)

1) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 30 minutes, the system of hypothesis would be:

Null hypothesis:
\mu = 30

Alternative hypothesis:
\mu \\eq 30

If we analyze the size for the sample is > 30 but 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=(\bar X-\mu_o)/((s)/(√(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".

2) Calculate the statistic

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


t=(30.8-30)/((1.5)/(√(32)))=3.017

3) P-value

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


df=n-1=32-1=31

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


p_v =2*P(t_((31))>3.017)=0.0051

4) 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 reject the null hypothesis, so we can conclude that the treu mean is different from 30 minutes at 1% of significance

User Eisa Adil
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