Question

How much time do you spend talking on your phone per day? Initial post by Wednesday at 11:59 PM: Guess the number of minutes you think you spend talking on the phone each day. This is your null hypothesis. Posting from Thursday to Sunday, 11:59 PM: Secure the data from your phone for the past month and conduct a one sample hypothesis test of the mean length of your phone calls per day. Be sure to show all your work by using excel and include your data set. Please check out the video if you need help on using excel to find the mean and sample standard deviation. Video on using excel Use a significance level of 0.05. You will have 30 data values showing the total of minutes you spend per day. (The easiest way to secure this data is to look at your "Recent" calls on your phone and list the total number of minutes that you have spent on the phone each day in the last 30 days. This will be the data that you will use.)

My guess is 60 mins per day, below is the data to utilize........

1	16
2	46
3	92
4	62
5	33
6	14
7	7
8	122
9	19
10	66
11	67
12	74
13	44
14	54
15	22
16	29
17	36
18	20
19	91
20	33
21	16
22	8
23	23
24	41
25	44
26	19
27	8
28	33
29	26
30	68

Answer 1

USING EXCEL'S FUNCTION ("=AVERAGE" and "=STDEV") we have

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS Ha: Minutes

Level Of significance=0.05

Rejection Region: Based on the information provided, the significance level is \alpha = 0.05α=0.05, and the critical value for a two-tailed test is tc​=2.045.

The rejection region for this two-tailed test is R={t:∣t∣>2.045}

Test Statistics

The t-statistic is computed as follows:

Decision about the null hypothesis: Since it is observed that ∣t∣=3.642>tc​=2.045, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.001, and since p=0.001<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is different than 60 min, at the 0.05 significance level.