169k views
4 votes
A random sample of n = 40 observations from a quantitative population produced a mean x = 2.2 and a standard deviation s = 0.29. Your research objective is to show that the population mean μ exceeds 2.1. Do the data provide sufficient evidence to conclude that μ > 2.1? (Use a 5% significance level.)

User Laszlowaty
by
5.2k points

1 Answer

2 votes

Answer:


t=(2.2-2.1)/((0.29)/(√(40)))=2.18


df=n-1=40-1=39

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


p_v =P(t_((39))>2.18)=0.0177

If we compare the p value and the significance level given
\alpha=0.05 we see that
p_v<\alpha so we can conclude that we have enough evidence to reject the null hypothesis, and then the population mes seems to be higher than 2.1 at 5% of significance

Explanation:

Data given and notation


\bar X=2.2 represent the sample mean


s=0.29 represent the sample standard deviation


n=40 sample size


\mu_o =2.1 represent the value that we want to test


\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 is higher than 2,1, the system of hypothesis would be:

Null hypothesis:
\mu \leq 2.1

Alternative hypothesis:
\mu > 2.1

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".

Calculate the statistic

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


t=(2.2-2.1)/((0.29)/(√(40)))=2.18

P-value

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


df=n-1=40-1=39

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


p_v =P(t_((39))>2.18)=0.0177

Conclusion

If we compare the p value and the significance level given
\alpha=0.05 we see that
p_v<\alpha so we can conclude that we have enough evidence to reject the null hypothesis, and then the population mes seems to be higher than 2.1 at 5% of significance

User Kevin Monk
by
4.7k points