Question

A random sample of 100 people from City A has an average IQ of 120 with a SD of 18. Independently of this, a random sample of 150 people from City B has an average IQ of 116 with a SD of 15. Are residents of City A smarter on average? Make a z-test of the difference in average IQs; the null hypothesis is that there is no difference.

Answer 1

Answer:

$z=\frac{(120-116)-0}{\sqrt{\frac{18^2}{100}+\frac{15^2}{150}}}}=1.837$

$p_v =P(z>1.837)=1-P(Z$  

Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the average IQ on city A is signficantly higher than city B at 5% of singificance.  

Step-by-step explanation:

$\bar X_{A}=120$ represent the mean for sample 1

$\bar X_{B}=116$ represent the mean for sample 2

$s_{A}=18$ represent the sample standard deviation for 1  

$s_{B}=15$ represent the sample standard deviation for 2  

$n_{A}=100$ sample size for the group 2  

$n_{B}=150$ sample size for the group 2  

$\alpha$ Significance level provided

z would represent the statistic (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if residents of City A smarter on average, the system of hypothesis would be:  

Null hypothesis:$\mu_{A}-\mu_{B}\leq 0$  

Alternative hypothesis:$\mu_{A} - \mu_{B}> 0$  

We don't have the population standard deviation's, but the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:  

$z=\frac{(\bar X_{A}-\bar X_{B})-\Delta}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:  

$z=\frac{(120-116)-0}{\sqrt{\frac{18^2}{100}+\frac{15^2}{150}}}}=1.837$

P value

Since is a one right tailed test the p value would be:  

$p_v =P(z>1.837)=1-P(Z$  

Comparing the p value with a significance level for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the average IQ on city A is signficantly higher than city B at 5% of singificance.