Respuesta :
Answer:
Yes, this provide enough evidence to show a difference in the proportion of households that own a desktop.
Step-by-step explanation:
We are given that National data indicates that 35% of households own a desktop computer.
In a random sample of 570 households, 40% owned a desktop computer.
Let p = population proportion of households who own a desktop computer
SO, Null Hypothesis, [tex]H_0[/tex] : p = 25% {means that 35% of households own a desktop computer}
Alternate Hypothesis, [tex]H_A[/tex] : p [tex]\neq[/tex] 25% {means that % of households who own a desktop computer is different from 35%}
The test statistics that will be used here is One-sample z proportion statistics;
T.S. = [tex]\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of 570 households who owned a desktop computer = 40%
n = sample of households = 570
So, test statistics = [tex]\frac{0.40-0.35}{{\sqrt{\frac{0.40(1-0.40)}{570} } } } }[/tex]
= 2.437
Since, in the question we are not given with the level of significance at which to test out hypothesis so we assume it to be 5%. Now at 5% significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics doesn't lies within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that % of households who own a desktop computer is different from 35%.
