In: Statistics and Probability
How much time do you spend talking on your phone per day? I would like you to guess the number of minutes. This is your null hypothesis. I then want you to 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. 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.)
The number of minuets guessed spent on the phone each day is 25 mins.
The minutes for each day spent on the phone for the past month are: 22, 15, 34, 58, 20, 15, 0, 0, 27, 63, 13, 22, 31, 23, 19, 15, 38, 57, 0, 9, 21, 61, 26, 28, 12, 24, 17, 27, 55, 16
How much time do you spend talking on your phone per day? I would like you to guess the number of minutes. This is your null hypothesis. I then want you to 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. 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.)
The number of minuets guessed spent on the phone each day is 25 mins.
The minutes for each day spent on the phone for the past month are: 22, 15, 34, 58, 20, 15, 0, 0, 27, 63, 13, 22, 31, 23, 19, 15, 38, 57, 0, 9, 21, 61, 26, 28, 12, 24, 17, 27, 55, 16
calculated t=0.1862
Table value of t with 29 DF at 0.05 level =2.0452
Rejection Region: Reject Ho if t < -2.0452 or t > 2.0452
Calculated t = 0.1862 , not in the rejection region
The null hypothesis is not rejected.
There is not enough evidence to reject the guess.
t Test for Hypothesis of the Mean |
|
Data |
|
Null Hypothesis m= |
25 |
Level of Significance |
0.05 |
Sample Size |
30 |
Sample Mean |
25.6 |
Sample Standard Deviation |
17.6510 |
Intermediate Calculations |
|
Standard Error of the Mean |
3.2226 |
Degrees of Freedom |
29 |
t Test Statistic |
0.1862 |
Two-Tail Test |
|
Lower Critical Value |
-2.0452 |
Upper Critical Value |
2.0452 |
p-Value |
0.8536 |
Do not reject the null hypothesis |
t Test for Hypothesis of the Mean |
|
Data |
|
Null Hypothesis m= |
25 |
Level of Significance |
0.05 |
Sample Size |
=COUNT(Sheet1!$A$1:$A$31) |
Sample Mean |
=AVERAGE(Sheet1!$A$1:$A$31) |
Sample Standard Deviation |
=STDEV.S(Sheet1!$A$1:$A$31) |
Intermediate Calculations |
|
Standard Error of the Mean |
=B8/SQRT(B6) |
Degrees of Freedom |
=B6-1 |
t Test Statistic |
=(B7-B4)/B11 |
Two-Tail Test |
|
Lower Critical Value |
=-(T.INV.2T(B5, B12)) |
Upper Critical Value |
=T.INV.2T(B5, B12) |
p-Value |
=T.DIST.2T(ABS(B13), B12) |
=IF(B18<$B$5, "Reject the null hypothesis", "Do not reject the null hypothesis") |