Question

Snapchat is interested in targeting a particular audience group with a new product(Spectacles)and needs to know about the average time a first-year college student spends on Snapchateach day. This way, they can decide whether or not it would be worthwhile to advertise Spectacles to students. Unfortunately, the Snapchatreport does not include specific information about first-year college students, so they don't have u for this population. They hire Serena to find more information.
She does some research online and finds that Website A claims that first-year college students spend on average 4 hours on Snapchatper day and Website B claims that first-year college students spend on average 3 hours on Snapchatper day. Since both sites made different claims, Serena decides to survey first-year students at De Anza to create a confidence interval to decide if the claims are reasonable. The following is a list of 30 responses (in hours):

2	1.5	6	3	0.75	2.5	1.25	0.25	0	0
2	2	1	1	3	4	5	2.2	3	6
6.5	5	1	1	1.5	3	5.2	6	4.5	6

1. Input the 30 observations into the list 1L of your calculator and use 1-Var-Stats to determine the sample mean X and the sample standard deviation .sRound your answers to 4 decimal places. Again, assume X is approximately normal.
X=_____________ s= ________________
2. Explain in one sentence why the t distribution must be used to create a confidence interval in this situation.
3. Using the information from problems 2 and 3, construct a 95% confidence interval for the population mean the number of hours spent on Snapchatby first-year college students. Round your answer to 4 decimal places. Show all of your work and interpret the confidence interval.
4. Are the claims made by Website A and Website B reasonable or unreasonable? Make your decision based on your 95% confidence interval from problem 4 and explain.
A. Website A’s claim of u=4 hours is reasonable/unreasonable(circle one) because...
B. Website B’s claim of u=3 hours is reasonable/unreasonable(circle one) because...

Answer 1

data	data - mean	(data - mean)2
2	-0.8717	0.75986089
1.5	-1.3717	1.88156089
6	3.1283	9.78626089
3	0.1283	0.01646089
0.75	-2.1217	4.50161089
2.5	-0.3717	0.13816089
1.25	-1.6217	2.62991089
0.25	-2.6217	6.87331089
0	-2.8717	8.24666089
0	-2.8717	8.24666089
2	-0.8717	0.75986089
2	-0.8717	0.75986089
1	-1.8717	3.50326089
1	-1.8717	3.50326089
3	0.1283	0.01646089
4	1.1283	1.27306089
5	2.1283	4.52966089
2.2	-0.6717	0.45118089
3	0.1283	0.01646089
6	3.1283	9.78626089
6.5	3.6283	13.16456089
5	2.1283	4.52966089
1	-1.8717	3.50326089
1	-1.8717	3.50326089
1.5	-1.3717	1.88156089
3	0.1283	0.01646089
5.2	2.3283	5.42098089
6	3.1283	9.78626089
4.5	1.6283	2.65136089
6	3.1283	9.78626089
Total = 86.15		121.9234

1.  

Sample standard deviation (s) =

  

2. In this context , t distribution will be used to find confidence interval because population standard deviation is unknown and sample size n ≤ 30 .

3.  We need to construct the 95\%95% confidence interval for the population mean μ.

The critical value for α=0.05 and df=n−1=29 degrees of freedom is t_c​=2.045. The corresponding confidence interval is computed as shown below:

4. Website B’s claim of u=3 hours is reasonable/unreasonable(circle one) because it fall within the 95% confidence interval.