In: Statistics and Probability
| Question 10: Perform a z-test to determine if the mean age is different than 60 years. (13 pts.) | |||||||||
| Data | |||||||||
| Null Hypothesis | 60.0 | ||||||||
| Level of Significance | 0.05 | ||||||||
| Population Standard Deviation | 5.0767272 | ||||||||
| Sample Size | 101 | ||||||||
| Sample Mean | 56.29703 | ||||||||
| Intermediate Calculations | |||||||||
| Standard Error of the Mean | 0.5052 | ||||||||
| Z Test Statistic | |||||||||
| Two-Tail Test | |||||||||
| Lower Critical Value | |||||||||
| Upper Critical Value | |||||||||
| p-Value | |||||||||
Please use excel and explain clearly with the formula used. I cannot put the data as it is too long but for your information, I have 101 data of one column. Use any data you want.
| Data | ||
| Null Hypothesis | 60 | |
| Level of significance | 0.05 | |
| Population Standard Deviation | 5.0767272 | |
| Sample Size | 101 | |
| sample Mean | 56.29703 |
| Intermediate Calculation | Formula Used | ||
| Standard Error of the mean | 0.5052 | = 5.0767272/SQRT(101) | |
| Z test Statistic | -7.32971101 | = (56.29703- 60)/0.5052 |

| Two tail Test | Formula Used | ||
| Lower Critical Value | -1.95996398 | = NORMINV(0.025,0,1) | |
| Upper Critical Value | 1.959963985 | = NORMINV(1-0.025,0,1) | |
| p value | 2.3065E-13 | = 2*NORMDIST(-7.3297,0,1,1) |
Last Column is formula used
(For p value we should multiply it by 2 as it is two tailed so we need to find p(z<-7.3297) and p(z> +7.3297) so due to symmetry of normal distribution, multiplication of 2 is valid)