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.

The degrees of freedom are given by:

Now we can calculate the p value with this probability:

  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.

Step-by-step explanation:

Information given

sample size  

represent the value to check

represent the significance level

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:  

Alternative hypothesis:  

The statistic is given by:

(1)  

For this case the statistic is given by:

The degrees of freedom are given by:

Now we can calculate the p value with this probability:

  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.

what

a_n = a_1 + (n - 1)d

the first term is 6.

the common difference d = ?

when you find d, replace n with 7 and d with the common difference to find your answer.