Answer:
We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.
Step-by-step explanation:
Jenna claims that the average time of texting at her larger high school is greater than 94 minutes per day.
From here we can see that we have to perform a hypothesis test about a sample mean. The null and alternate hypothesis will be:
Null Hypothesis: ![\mu \leq 94](https://tex.z-dn.net/?f=%5Cmu%20%5Cleq%20%2094)
Alternate Hypothesis: ![\mu > 94](https://tex.z-dn.net/?f=%5Cmu%20%3E%2094)
Jenna collected data from a sample of 32 students. So, sample size will be:
Sample Size = n = 32
Sample Mean = x = 96.5
Sample Standard Deviation = s = 6.3
We have to perform a hypothesis test, to test Jenna's claim. Since, the value of Population Standard Deviation is unknown and the value of Sample Standard Deviation is known, we will use One Sample t-test in this case.
The formula to calculate the test statistic is:
![t=\frac{x-\mu}{\frac{s}{\sqrt{n}}}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Cfrac%7Bs%7D%7B%5Csqrt%7Bn%7D%7D%7D)
Using the values, we get:
![t=\frac{96.5-94}{\frac{6.3}{\sqrt{32} } }=2.245](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B96.5-94%7D%7B%5Cfrac%7B6.3%7D%7B%5Csqrt%7B32%7D%20%7D%20%7D%3D2.245)
The degrees of freedom will be:
df = n - 1 = 32 - 1 = 31
We have to convert the t-score 2.245 with 31 degrees of freedom to its equivalent p-value. From t-table this value comes out to be:
p-value = 0.0160
The significance level is:
![\alpha =0.05](https://tex.z-dn.net/?f=%5Calpha%20%3D0.05)
Since, the p-value is lesser than the level of significance, we reject the Null Hypothesis.
Conclusion:
We have sufficient evidence to support the claim that the average for the students at Jenna's large high school is greater than 94 minutes.