Respuesta :
Answer:
There is no significant difference between the two regions population proportion of people who go for online shopping
Step-by-step explanation:
We set the null hypothesis as H₀ : p₁ = p₂
Our alternative hypothesis is then Hₐ : p₁ ≠ p₂
Here we have the test statistic as follows;
[tex]Z_0 =\frac{p_1 - p_2}{\sqrt{\frac{p_1q_1}{n_1} +\frac{p_2q_2}{n_2} } }[/tex]
Where:
p₁ = 16%
p ₂= 24%
n₁ = 100 people
n₂ = 100 people
q = 1 - p
Therefore by plugging the values we get,
[tex]Z_0 =\frac{0.16 - 0.24}{\sqrt{\frac{0.16\times 0.84}{100} +\frac{0.24\times 0.76}{100} } } = -1.42134[/tex]
From the z relations the probability is found to be P(Z = -1.42) = 0.1573
Therefore, as p > α at 0.05, we fail to reject the null hypothesis, and therefore based on the statistical evidence, there is no significant difference between the two regions population proportion of people who go for online shopping
The proportion of population doing online shopping from both region aren't significantly different.
Given that:
For region A:
- p₁ = 16% = 0.16 probability
- n₁ = 100 people
For region B:
- p ₂= 24% = 0.24 probability
- n₂ = 100 people
Laying down the hypothesis:
The null Hypothesis [tex]H_0: p_1 = p_2[/tex]
Alternative Hypothesis [tex]H_1: p_1 \neq p_2[/tex]
The test statistic can be calculated as:
[tex]Z_0 = \dfrac{p_1 - p_2}{\sqrt{\dfrac{p_1(1-p_1)}{n_1} + \dfrac{p_2(1-p_2)}{n_2}}}\\[/tex]
Putting values will give:
[tex]Z_0 = -1.4213[/tex]
From the p values of Z score, the probability is calculated:
[tex]P(Z = -1.42) = 0.1573[/tex]
Thus, as the probability is bigger than the value of alpha at significance level 0.05, thus we can't reject null hypothesis.
The proportion of population doing online shopping from both region aren't significantly different.
Learn more about null and alternate hypothesis:
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