Question

Suppose you carry out a significance test of H0: μ = 3.5 versus Ha: μ < 3.5 based on sample size n = 17 and obtain t = –3.4. Find the p-value for this test. What conclusion can you draw at the 5% significance level? Explain. (4 points)

Group of answer choices


The p-value is 0.4982. We reject H0 at the 5% significance level because the p-value 0.4982 is greater than 0.05.


The p-value is 0.0018. We reject H0 at the 5% significance level because the p-value 0.0018 is less than 0.05.


The p-value is 0.5018. We fail to reject H0 at the 5% significance level because the p-value 0.5018 is greater than 0.05.


The p-value is 0.4982. We fail to reject H0 at the 5% significance level because the p-value 0.4982 is greater than 0.05.


The p-value is 0.0018. We fail to reject H0 at the 5% significance level because the p-value 0.0018 is less than 0.05.

Answer 1

The p-value is 0.0018. We reject H0 at the 5% significance level because the p-value 0.0018 is less than 0.05.

Explanation:

I took the quiz and got it right

Answer 2

`t =-3.4`

The degrees of freedom are given by:

`df = n-1= 17-1 =16`

Now we can calculate the p value with this probability:

`p_v =P(t_(16)<-3.4)=0.0018` And for this case the p value is lower than the significance level and the best conclusion would be:

The p-value is 0.0018. We reject H0 at the 5% significance level because the p-value 0.0018 is less than 0.05.

Explanation:

Information given

`n=17` sample size

`\mu_o =3.5` represent the value to check

`\alpha=0.05` represent the significance level

`p_v` represent the p value

Hypothesis to verify

We want to verify if the true mean for this case is lower than 3.5, the system of hypothesis would be:

Null hypothesis:
`\mu \geq 3.5`

Alternative hypothesis:
`\mu < 3.5`

The statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

For this case the statistic is given by:

`t =-3.4`

The degrees of freedom are given by:

`df = n-1= 17-1 =16`

Now we can calculate the p value with this probability:

`p_v =P(t_(16)<-3.4)=0.0018` And for this case the p value is lower than the significance level and the best conclusion would be:

The p-value is 0.0018. We reject H0 at the 5% significance level because the p-value 0.0018 is less than 0.05.