A group of students wanted to investigate the claim that the average number of text messages sent yesterday by students in their school was greater than 100. They asked each student in a random sample of 50 students how many text messages he or she sent yesterday. An appropriate t-test was conducted and resulted in a p-value of 0.0853. Assuming the conditions for the t-test were met, which of the following is an appropriate conclusion?

A. Because p-value < 0.10, at the 10% significance level, it can be concluded that the mean number of test messages sent yesterday by the students is less than 100.

B. Because p-value < 0.10, at the 10% significance level, it cannot be concluded that the mean number of test messages sent yesterday by the students is greater than 100.

C. Because p-value > 0.05, at the 5% significance level, it can be concluded that the mean number of test messages sent yesterday by the students is greater than 100.

D. Because p-value > 0.05, at the 5% significance level, it can be concluded that the mean number of test messages sent yesterday by the students is less than 100.

E. Because p-value > 0.05, at the 5% significance level, it cannot be concluded that the mean number of test messages sent yesterday by the students is greater than 100.

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis. And the best option would be:

E. Because p-value > 0.05, at the 5% significance level, it cannot be concluded that the mean number of test messages sent yesterday by the students is greater than 100.

Step-by-step explanation:

Data given and notation

[tex]\bar X[/tex] represent the sample mean

[tex]s[/tex] represent the sample standard deviation

[tex]n=50[/tex] sample size

[tex]\mu_o =100[/tex] represent the value that we want to test

[tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

[tex]p_v[/tex] 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 greater than 100, the system of hypothesis are :

Null hypothesis:[tex]\mu \leq 100[/tex]

Alternative hypothesis:[tex]\mu > 100[/tex]

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:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (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

The statistic calulated for this case is [tex] t_{calc}[/tex]

P-value

We can find the degrees of freedom are given by:

[tex] df = n+1=50-1=49[/tex]

Since is a one-side right tailed test the p value would given by:

[tex]p_v =P(t_{49}>t_{calc})=0.0853[/tex]

Conclusion

If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis. And the best option would be:

E. Because p-value > 0.05, at the 5% significance level, it cannot be concluded that the mean number of test messages sent yesterday by the students is greater than 100.