The average age of a group of people is 63
The standard deviation is 0.05.
Is there enough information given to support this claim?
If 'yes', then what is the 'T' value?
Any help would be great, thanks!!
2007-07-23 11:05:08 · 3 answers · asked by wow i need help 1 in Science & Mathematics ➔ Mathematics
No, there is not (nearly) enough information. Your question has to answer the following:
Are the mean and standard deviation that you give, the population parameters or the sample parameters?
What is the sample size?
What is your null hypothesis?
What is the α ? I have a feeling that the 0.05 you give is the α risk and not the standard deviation.
This information is required to calculate the t-value.
2007-07-23 11:16:50 · answer #1 · answered by cvandy2 6 · 0⤊ 0⤋
All the answers thus far are correct. But you know what, you can make a WAG, which is sometimes all you really need to support a decision. The WAG is based on something called the Three Point Estimate. It consists of the following approximations:
Average (mean) = (UB + 4 ML + LB)/6 = 63
Std Dev = (UB - LB)/6 = .05
UB and LB are upper and lower bounds in your data. ML is the most likely data value (the mode).
So, here are some of the WAGs you can make.
2/3 of the ages will lie between 63 - .05 and 63 + .05 years.
2.5% of the ages will be above 63 + 2*.05 and 2.5% below 63 - 2*.05 years. Thus, 95% of the ages will lie inbetween.
And virtually none of the ages will be above UB = 63 + 3*.05 or below LB = 63 - 3*.05.
And you can further WAG that the mode = average.
All these WAGs are derived from the supposition your data are Normally distributed.
2007-07-23 18:38:58 · answer #2 · answered by oldprof 7 · 0⤊ 0⤋
How large is the group? How big is the sample? I don't think you have enough information to make any claims.
2007-07-23 18:10:39 · answer #3 · answered by Philo 7 · 0⤊ 1⤋