How to calculate Mean and standard deviation without sample size in Minitab

[Artie21]

Oops, i made a small error, the title was meant to say 'How to calculate Mean and standard deviation WITHOUT sample size in Minitab?

Hi everyone? How are you all?

I'm a new member and I'm running into a bit of a problem trying to figure out how to calculate the mean and standard deviation without a sample size.

I was given a problem:

A sample of 85 puppies under the age of 25 months were tested and of these 22 admitted for 'hyperactivity'. Is there sufficient evidence at a = 0.05 to show that the unknown population proportion exceeds 30%?

I've always been given the sample mean and standard deviation for a problem to calculate the test statistics in Minitab but this time, the teacher left that information out

I can't use the Z-test or the T-test because there isn't enough information to compute the test, the mean and standard deviation is missing

From my understanding, the Null hypothesis μ = 30% Vs Alternative hypothesis μ > 30%.

When a = 0.05 , the confidence interval would be at 95% and the Z-score/value is 1.645 to the right.

How do i compute the P-value and Z-value without the mean and standard deviation using Minitab??

How do i solve the problem at all!!

Thanks and i appreciate for you help

 

[Miner]

Try a 1-proportion test.

 

[Artie21]

Hi Miner, thanks for the reply

I have looked into the 1-proportion test but in-order to compute the One-Sample Proportion test, i would need to input the 'number of trials' and the 'number of events' which are the values that i don't have or know.

All i got from the problem are the sample size of 85 puppies and 22 of them are Hyperactive and I need to show if the unknown population proportion exceeds 30%. I would need to use the Null Hypothesis Vs Alternative Hypothesis test - either Z-sample or T-Sample test to show the rejection region and critical value. BUT I can't run the Z-sample test or the T-sample test without the Mean and Standard Deviation of the sampled population

Or should I just say there isn't enough sufficient data to compute the test?

This is why i'm asking is there anyway in Minitab that allows me to calculate the Mean and standard Deviation from the sample of '85' puppies.

 

[Miner]

Number of trials = 85
Number of events = 22

 

[Bev D]

Miner is correct (as always)
think about it - which is what your instructor wants you to do

 

[Artie21]

Thanks guys for your replies.

Yes, I followed Miner's advice/instruction again and got the 1-Proportion test to compute in minitab

I'm not good with maths or Statistic calculations so I really appreciate your help!

Thanks guys for your time! You guys are the best!