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Answer:
If we compare the p value obtained and the significance level given we have
so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 2% of significance the proportion of arrests that were not prosecuted is not significanlty less than 0.31.
Step-by-step explanation:
1) Data given and notation
n=500 represent the random sample taken
X represent the number of arrests that were not prosecuted.
estimated proportion of arrests that were not prosecuted
is the value that we want to test
represent the significance level
Confidence=98% or 0.98
z would represent the statistic (variable of interest)
represent the p value (variable of interest)
2) Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.31.:
Null hypothesis:
Alternative hypothesis:
When we conduct a proportion test we need to use the z statisitc, and the is given by:
(1)
The One-Sample Proportion Test is used to assess whether a population proportion is significantly different from a hypothesized value
.
3) Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
4) Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided . The next step would be calculate the p value for this test.
Since is a left tailed test the p value would be:
If we compare the p value obtained and the significance level given we have
so we can conclude that we have enough evidence to FAIL reject the null hypothesis, and we can said that at 2% of significance the proportion of arrests that were not prosecuted is not significanlty less than 0.31.
Answer:
P(Y ≥ 15) = 0.763
Step-by-step explanation:
Given that:
Mean =135
standard deviation = 12
sample size n = 50
sample mean = 140
Suppose X is the random variable that follows a normal distribution which represents the weekly supermarket expenses
Then,
The probability that X is greater than 140 is :
P(X>140) = 1 - P(X ≤ 140)
From z tables,
Similarly, let consider Y to be the variable that follows a binomial distribution of the no of household whose expense is greater than $140
Then;
∴
P(Y ≥ 15) = 1- P(Y< 15)
P(Y ≥ 15) = 1 - ( P(Y=0) + P(Y=1) + P(Y=2) + ... + P(Y=14) )
P(Y ≥ 15) = 0.763