Question

Cody is an environmental scientist inquiring about the percentage of deer ticks in a certain national park that are infected with Lyme disease. He suggests that more than 28% of the deer ticks in the park are infected with the disease. Cody randomly selects deer ticks from various spots throughout the park and tests each one for the presence of the disease. Of the 141 deer ticks Cody selected, 43 were infected with Lyme disease. Are all of the conditions for this hypothesis test met, and if so, what are the null and alternative hypotheses for this hypothesis test?

Answer 1

Cody's hypothesis test to determine the percentage of deer ticks infected with Lyme disease meets all the necessary conditions: random sampling, normality, and independence. Therefore, the null hypothesis is that the proportion of infected ticks is 28% or less, and the alternative hypothesis is that it's greater than 28%.

Step-by-step explanation:

Cody is conducting a hypothesis test to determine if the true proportion of deer ticks in a national park infected with Lyme disease is greater than 28%. To do this, the following conditions must be met:

• Random sampling: Cody randomly selects deer ticks from various locations within the park. This condition appears to be met based on the information provided.
• Normality: The sample size should be large enough to approximate a normal distribution of the sample proportion. In this case, both np and n(1-p) should be greater than 5 for the approximation to hold. For Cody's sample: np = 141 × 0.28 = 39.48 and n(1-p) = 141 × 0.72 = 101.52, both of which are greater than 5.
• Independence: Each tick's chance of having Lyme disease should be independent of the others. This is often assumed if the population from which the sample is drawn is at least 10 times larger than the sample (which is likely true here).

The null hypothesis (H0) for this test would be that the true proportion of infected deer ticks is 28% or less, while the alternative hypothesis (H1) is that the true proportion is greater than 28%.

Answer 2

In order to apply the z proportion test we need to verify the following conditions:

1) Randomization we need a random sample.

2) 10 % condition we need that the sample size would be lower than 10% of the population size

3) We need to satisfy the normalyti condition:

n⋅p≥10 and n⋅(1−p)≥10

And for this case we satisfy all the conditions

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.28, so the correct system of hypothesis are:

Null hypothesis:
`p \leq 0.28`

Alternative hypothesis:
`p > 0.28`

Step-by-step explanation:

Data given and notation

n=141 represent the random sample taken

X=43 represent the adults that were infected with Lyme disease

`\hat p=(43)/(141)=0.305` estimated proportion of adults infected with Lyme disease

`p_o=0.28` is the value that we want to test

`\alpha=` represent the significance level

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

Solution to the problem

In order to apply the z proportion test we need to verify the following conditions:

1) Randomization we need a random sample.

2) 10 % condition we need that the sample size would be lower than 10% of the population size

3) We need to satisfy the normalyti condition:

n⋅p≥10 and n⋅(1−p)≥10

And for this case we satisfy all the conditions

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.28, so the correct system of hypothesis are:

Null hypothesis:
`p \leq 0.28`

Alternative hypothesis:
`p > 0.28`