Respuesta :
Answer:
Step-by-step explanation:
If p is the sample proportion of thefts and frauds for identify then population proportion is 23%
P = 0.23
[tex]\sigma = \sqrt{\frac{PQ}{n} } \\=0.0111[/tex]
Sample proportion p = 0.224
p diff = -0.076
Test statistic Z = p diff/se = -0.339
p value = 0.738
SInce p >alpha of 5% we accept H0
There is no statistical evidence to prove that this data provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%
This data not enough provide the evidence to show that Alaska had a lower proportion of identity theft than 23%
Explanation:
According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. In that year, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints ("Consumer fraud and," 2008). Does this data provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%? Test at the 5% level.
The random variable x = number of complaints from identity theft in Alaska = 321
The parameter of interest p = proportion of complaints from identity theft in Alaska = 0.23
The hypotheses for this experiment are:
null H0: P = 0.23
alternative HA: P < 0.23
The level of significance α = 0.05
A simple random sample of the category of 1432 complaints of identity theft in Alaska were gathered. The study says out of all complaints that year, the data maybe filed at random period. This assumption maybe met, but the data cannot be real.
There are 1432 complaints in this sample. The reason for the complaint does not affect the next complaint. There can only be two outcomes that can be used to this type of scenario, the complaint was for identity theft or it was not. The chance that one complaint was for identity theft does not change. Therefore the conditions for the binomial distribution are satisfied.
In this case
p = 0.23
n = 1432
np = 1432*0.23 = 329.36 ≥ 5
nq = 1432*(1 – 0.23) = 1102.64 ≥ 5.
the sampling distribution is a normal distribution; this means we will use a z-test.
The test statistic
[tex]y = \frac{(x/n)-p}{\sqrt{ \frac{p(1-p) }{n} } } = \frac{0.2242-0.23}{\sqrt{ \frac{0.23(1-0.23) }{1432} } } = -0.522[/tex]
The p-value associated with this problem is given by:
[tex]= NORM.S.DIST(z, cumulative) \\= NORM.S.DIST (-0.522, TRUE) = 0.2998[/tex]
Since the p-value is greater than the level of significance such as [tex][p-value = 0.2998] > [\alpha = 0.05][/tex] we fail to reject
Therefore it is not enough evidence to determine that the proportion of complaints due to identity theft in Alaska is less than [tex]23%.[/tex]%
Learn more about the random variable https://brainly.com/question/13603146
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