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The mean IQ for a population is 100, and the standard deviation is 15. Girard’s IQ is 142. Given the percentage of people with IQs as high as or higher than Girard’s in this population, does it seem like Girard’s IQ is high enough to reject the null hypothesis? Please justify?

Aug 16th, 2015

The basic z score formula for finding a z-score for a sample is:
z = x – μ / σ

This is exactly the same formula as z = x – μ / σ, except that Xbar (the sample mean) is used instead of μ (the population mean) and s (the sample standard deviation) is used instead of σ (the population standard deviation). The steps for solving it are exactly the same.

When you have multiple samples and want to describe the standard deviation of those sample means, you would use this z score formula:
z = x – μ / (σ / √n)
This z score will tell you how many standard errors there are between the sample mean and the population mean.
Sample problem: In general, the mean height of women is 65″ with a standard deviation of 3.5″. What is the probability of finding a random sample of 50 women with a mean height of 70″, assuming the heights are normally distributed?

z = x – μ / (σ / √n)
= 160 – 142 / (15/√100) = 18 / 150 = 0.12

We know that 99% of values fall within 3 standard deviations from the mean in a normal probability distribution. Therefore, Girard’s IQ is high enough to reject the null hypothesis.

Aug 16th, 2015

This can also be done using  f-value and p-value

for further details you can also read this

http://www.statisticshowto.com/f-value-one-way-anova-reject-null-hypotheses/

Aug 16th, 2015

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Aug 16th, 2015
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Aug 16th, 2015
Oct 17th, 2017
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