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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

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

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