Complete #2 for me and show proof , will give brainliest to whose right .
1 answer:
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Answer:
B=12.2
C=5.4 is your answer
Answer:
Comparing the p value with a significance level for example we see that
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:
represent the mean for sample 1
represent the mean for sample 2
represent the sample standard deviation for 1
represent the sample standard deviation for 2
sample size for the group 2
sample size for the group 2
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:
Alternative hypothesis:
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:
(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:
P value
Since is a one right tailed test the p value would be:
Comparing the p value with a significance level for example we see that
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.