Question

According to a certain study, of all the households in a city that owned at least one laptop computer, 85% has two or more laptop computers. An internet provider planning to promote a new internet service package found that out of the 400 randomly selected households visited, 310 had two or more laptop computers. At the 0.05 level of significance, is there sufficient evidence to conclude that the proportion is less than what was reported? Express numerical answers in two decimals only.

1. What is the sample proportion?
2. What is the null hypothesis?
3. What is the alternative hypothesis?
4. Is the test left tailed or right tailed?
5. What is the critical value at 0.05 level?
6. What is the computed value of the test statistic?
7. What is the decision and conclusion?

Answer 1

Asnwer:

population proportion ( p ) = 85% = 0.85

sample size ( n ) = 400, x= 310

significance level ( $\alpha$ ) = 0.05

1)

$\hat p=\frac{x}{n}=\frac{310}{400}=0.775$

sample proportion = 0.775

2)

The null hypothesis is

Ho: p = 0.85

3)

alternative hypothesis is
H1: p < 0.85

4)

Test is left tailed

5)

now calculate z critical value for left tailed test with  $\alpha$ = 0.05

using normal z table we get

Critical value = -1.65

6)

formula for test statistics is

P-p p*(1-P)

$z =\frac{0.775 - 0.85}{\sqrt{\frac{0.775 * ( 1 - 0.775)}{400}}}$

z = -4.20

test statistic (z) = -4.20

c)

decision rule is

Reject Ho if ( test statistics ) $\leq$ ( critical value)

here, ( test statistic= -4.20 ) < ( Critical value = -1.65 )

Hence, we can say,

Null hypothesis is rejected.

Therefore there is enough sufficient evidence to conclude that the proportion is less than what was reported at 0.05 level of significance.