Question

The pesticide diazinon is in common use to treat infestations of the German cockroach, Blattella germanica. A study investigated the persistence of this pesticide on various types of surfaces. Researchers applied a 0.5% emulsion of diazinon to glass and plasterboard. After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On the glass, 18 cockroaches died, while on plasterboard, 25 died. Is there convincing evidence that the difference in the proportion of cockroaches that died on each surface is significant from the data given?

Answer 1

To determine if there is convincing evidence of a significant difference in the proportion of cockroaches that died on each surface, a two-sample proportion test can be performed using the given data. Applying the test formula yields a test statistic that does not exceed the critical value, indicating that we fail to reject the null hypothesis. Therefore, there is no convincing evidence to support a significant difference.

Step-by-step explanation:

To determine if there is convincing evidence that the difference in the proportion of cockroaches that died on each surface is significant, we can perform a hypothesis test. The null hypothesis (H0) is that there is no significant difference in the proportion of cockroaches that died on glass and plasterboard. The alternative hypothesis (Ha) is that there is a significant difference in the proportions.

We can use a two-sample proportion test to compare the proportions of cockroaches that died on each surface. The test statistic is calculated as:

test statistic = (p1 - p2) / sqrt( p * (1-p) * (1/n1 + 1/n2) )

Where p1 and p2 are the proportions of cockroaches that died on glass and plasterboard respectively, n1 and n2 are the sample sizes, and p is the pooled proportion. The pooled proportion is calculated as:

p = (x1 + x2) / (n1 + n2)

Using the given data, we have p1 = 18/36 = 0.5, p2 = 25/36 ~ 0.69, n1 = 36, and n2 = 36. Substituting these values into the test statistic formula gives:

test statistic = (0.5 - 0.69) / sqrt(0.595 * 0.405 * (1/36 + 1/36)) ≈ -1.37

Using a significance level of 0.05 (or 5%), we can compare the test statistic to the critical value for a two-tailed test. For a two-tailed test with a significance level of 0.05, the critical value is approximately ±1.96. Since the test statistic (-1.37) is not greater than -1.96 or less than -1.96, we fail to reject the null hypothesis. This means that there is not convincing evidence to suggest a significant difference in the proportion of cockroaches that died on glass and plasterboard based on the given data.

Answer 2

We conclude that there is no difference in the proportion of cockroaches that died on each surface.

Step-by-step explanation:

We are given that a study investigated the persistence of this pesticide on various types of surfaces.

After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On the glass, 18 cockroaches died, while on plasterboard, 25 died.

Let
`p_1` = proportion of cockroaches that died on glass surface.

`p_2` = proportion of cockroaches that died on plasterboard surface.

So, Null Hypothesis,
`H_0` :
`p_1-p_2` = 0 {means that there is no difference in the proportion of cockroaches that died on each surface}

Alternate Hypothesis,
`H_A` :
`p_1-p_2\\eq` 0 {means that there is a significant difference in the proportion of cockroaches that died on each surface}

The test statistics that would be used here Two-sample z proportion statistics;

T.S. =
`\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{(\hat p_1(1-\hat p_1))/(n_1)+(\hat p_2(1-\hat p_2))/(n_2) } }` ~ N(0,1)

where,
`\hat p_1` = sample proportion of cockroaches that died on glass surface =
`(18)/(36)` = 0.50

`\hat p_2` = sample proportion of cockroaches that died on plasterboard surface =
`(25)/(36)` = 0.694

`n_1` = sample of cockroaches on glass surface = 36

`n_2` = sample of cockroaches on plasterboard surface = 36

So, test statistics =
`\frac{(0.50-0.694)-(0)}{\sqrt{(0.50(1-0.50))/(36)+(0.694(1-0.694))/(36) } }`

= -1.712

The value of z test statistics is -1.712.

Since, in the question we are not given with the level of significance 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 lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference in the proportion of cockroaches that died on each surface.