Question

National data indicates that​ 35% of households own a desktop computer. In a random sample of 570​ households, 40% owned a desktop computer. Does this provide enough evidence to show a difference in the proportion of households that own a​ desktop? Identify the appropriate null and alternative hypotheses.

Answer 1

Yes, this provide enough evidence to show a difference in the proportion of households that own a​ desktop.

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,
`H_0` : p = 25% {means that 35% of households own a desktop computer}

Alternate Hypothesis,
`H_A` : p
`\\eq` 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. =
`\frac{\hat p-p}{{\sqrt{(\hat p(1-\hat p))/(n) } } } }` ~ N(0,1)

where,
`\hat p` = sample proportion of 570​ households who owned a desktop computer = 40%

n = sample of households = 570

So, test statistics =
`\frac{0.40-0.35}{{\sqrt{(0.40(1-0.40))/(570) } } } }`

= 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%.