Question

A recent study reported that high school students spend an average of 94 minutes per day texting. Jenna claims that the average for the students at her large high school is greater than 94 minutes. She will conduct a study to investigate this claim.

(b) Based on a sample of 32 students, Jenna calculated a sample mean of 96.5 minutes and a sample standard deviation of 6.3 minutes. Assume all conditions for inference are met. At the significance level of α=0.05, do the data provide convincing statistical evidence to support Jenna’s claim? Complete an appropriate inference procedure to support your answer.

Answer 1

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: $\mu \leq 94$

Alternate Hypothesis: $\mu > 94$

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:

$t=\frac{x-\mu}{\frac{s}{\sqrt{n}}}$

Using the values, we get:

$t=\frac{96.5-94}{\frac{6.3}{\sqrt{32} } }=2.245$

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:

$\alpha =0.05$

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.

Answer 2

This p value the significant level  0.05 is given which is higher then the obtained p value. Thus we can reject the null hypothesis. Thus we have convincing statistical evidence to support Jenna’s claim.

Given-

Jenna claims that the average for the students at her large high school is greater than 94 minutes. The null hypothesis is  $\mu\leq 94$  and alternate hypothesis  $\mu >94$.

Now given that,

The value of sample size $n$ is 32, sample means $x$ is 96.5 and the sample standard deviation $s$ is 6.3.

Here one-sample t-test will be used for the as sample standard deviation is given. The formula for the test statics is,

$t=\dfrac{x-\mu}{\dfrac{s}{\sqrt{n}} }$

$t=\dfrac{96.5-94}{\dfrac{6.3}{\sqrt{32}} }$

$t=2.245$

Now the degrees of freedom is n-1. thus,

$D_f=n-1$

$D_f=32-1=31$

Now from the t table the value of t score 2.245 having degree 31 is,

p value$=0.0160$

For this p value the significant level  0.05 is given which is higher then the obtained p value. Thus we can reject the null hypothesis. Thus we have convincing statistical evidence to support Jenna’s claim.

For more about the p value follow the link below-

https://brainly.com/question/14723549