Question

In a study of pet owners, it was reported that \(24 \%\) celebrate their pet's birthday (Pet Statistics, Bissell Homecare, Inc., 2010 ). Suppose that this estimate was from a random sample of 200 pet owners. Is it reasonable to conclude that the proportion of all pet owners who celebrate their pet's birthday is less than \(0.25 ?\) Use what you know about the sampling distribution of \(\hat{p}\) to support your answer.

Short answer

Given all the data and the calculation above, you will get a z-score and its corresponding P-value. If the P-value is lower than 0.05, which is the common threshold in most scientific fields, it is reasonable to conclude that less than 25% of the pet owners celebrate their pet's birthdays.

Step-by-step solution

Define the hypothesis

To determine if the proportion of pet owners who celebrate their pet's birthday is less than 0.25, you should set up the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). Here, the null hypothesis would be that the proportion (\(p\)) of all pet owners who celebrate their pet's birthday is equal to 0.25 (\(p = 0.25\)). The alternative hypothesis is that the proportion of all pet owners who celebrate their pet's birthday is less than 0.25 (\(p < 0.25\)).

Calculate the sample proportion

Our sample proportion (\(\hat{p}\)) is the number of successes (pet owners who celebrate their pet's birthday) divided by the number of trials (the total number of pet owners). This would be 0.24, as given in the problem.

Understand the sampling distribution of \(\hat{p}\)

According to the Central Limit Theorem, if the sample size is large enough (usually if n>30), the sampling distribution of \(\hat{p}\) will be approximately normally distributed. The mean of this distribution is equal to the population proportion (\(p\)), and the standard deviation is \(\sqrt{{p(1-p)}/{n}}\). In this case, the sample size is 200, which is large enough, so we can safely apply the Central Limit Theorem.

Standard deviation and z-score

Our sample size (\(n\)) is 200 and from the null hypothesis, \(p = 0.25\). Hence the standard deviation can be calculated as \(\sqrt{{(0.25*0.75)}/{200}}\). The z-score, which is the number of standard deviations our sample proportion (\(\hat{p}\)) is away from the mean under the null hypothesis, can be calculated as \((\hat{p} - p) / standard deviation\).

Decision

With the calculated z-score, you could look up the P-value (probability that a value is less than your sample mean) from a standard normal table. If P-value is low (typically less than 0.05), the null hypothesis is rejected, meaning it's reasonable to conclude that the proportion of pet owners who celebrate their pet's birthday is less than 0.25.