Question

Bobby took a random sample of 1000 trees to test for bug infestation. He found that there were 112 infested trees.

Does this sample provide significant evidence that more than 10% of trees are infested? Use a significance level of α = 0.01.

Find a 95% confidence interval for the proportion of infested trees.

Answer 1

Solution :

Given that ,

n = 1000

x = 112

The null and alternative hypothesis is

H₀ : p = 0.10

H_a : p > 0.10

This is the right tailed test .

$\hat p$ = x / n = 112 / 1000 = 0.112

P₀ = 10% = 0.10

1 - P₀ = 1 - 0.10 = 0.9

Test statistic = z

= $\hat p$ - P₀ / [$\sqrt$P_{0 *} (1 - P₀ ) / n]

= 0.112 - 0.10 / [$\sqrt$0.10 (1 - 0.10) / 1000]

= 1.265

The test statistic = 1.265

P-value = 0.1030

$\alpha$ = 0.01

0.1030 > 0.01

P-value > $\alpha$

Fail to reject the null hypothesis .

There is not sufficient evidence to test the claim .

Given that,

n = 1000

x = 112

Point estimate = sample proportion = $\hat p$ = x / n = 112 / 1000= 0.112

1 - $\hat p$ = 1 - 0.112 = 0.888

At 95% confidence level

$\alpha$ = 1 - 95%

$\alpha$ =1 - 0.95 =0.05

$\alpha$/2 = 0.025

Z$\alpha$/2 = Z_{0.025 = 1.960}

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.96 ($\sqrt$((0.112 * 0.888)) / 1000)

= 0.020

A 95% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.112 - 0.020 < p < 0.112 + 0.020

0.092 < p < 0.132

( 0.092 , 0.132 )

The 95% confidence interval for the proportion p is : ( 0.092 , 0.132 )