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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 15

Sloan [31]

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.

5 0

3 years ago

Flying against the wind, an airplane travels 4340 kilometers in 7 hours. Flying with the wind, the same plane travels 3120 kilom

Over [174]

The rate of the wind is given to be 210 the rate of the plane in mid air is given as 830

<h3>Howe to solve for the rate of the plane and wind</h3>

In 7 hours the kilometer traveled = 4340

P + W = 4340/7

= 620

P-W = 3120/3 hours

= 1040

2p = 1040 + 620