Question

The Wechsler Adult Intelligence Scale (IQ test) is constructed so that Full Scale IQ scores follow a normal distribution, with a mean of 100, and a standard deviation of 15. The mayor of Smart Town believes the county’s residents are smarter than the national average and wants to use it (the intelligence of the residents) as a marketing tool to bring industries to the area. A researcher is hired to conduct a study to determine whether Smart County’s residents have, on average, higher Full Scale IQs than the population. A random sample of 100 people from Smart County were given the IQ test and were found to have an average Full Scale IQ of 105. After analyzing the data to determine whether the mean Full Scale IQ score of the Smart County residents is higher than the national average, the p-value of 0.0004 was obtained. Using a 0.05 significance level, what conclusion can be drawn from the data?

Answer 1

Explanation:

Hello!

The variable "X: Full scale IQ score of an adult" follows a normal distribution with mean μ= 100 and standard deviation σ= 15

The claim is that the county's residents are smarter than the national average, symbolically: μ > 100

To test this a researcher took a sample of n=100 residents that took the test and recorded an average of X[bar]= 105

The hypotheses are:

H₀: μ ≤ 100

H₁: μ > 100

α: 0.05

p-value: 0.0004

Using the p-value approach, the decision rule is as follows:

p-value ≤ α, reject the null hypothesis.

p-value > α, do not reject the null hypothesis.

The p-value is less than the significance level so the decision is to reject the null hypothesis.

At a 5% significance level, you can conclude that the average full-scale IQ score of the Smart County residents is higher than the national average.

I hope this helps!