# Large sample proportion hypothesis testing

The issue:

A ketchup manufacturer is in the process of deciding whether to produce a new extra-spicy brand. The company's marketing-research department used a national telephone survey of 6,000 households and found that the extra-spicy ketchup would be purchased by 335 of them. A much more extensive study made 2 years ago showed that 5 percent of the households would purchase the brand then. At a 2 percent significance level, should the company conclude that there is an increased interest in the extra-spicy flavor?

My solution:

• Specify the hypothesis

$H_0:P=5\%$, $\; \; \; H_1 : P > 5 \%$,

it is an upper tailed test.

• Assume $H_0$ be true and desume population information,

• Compute the probability that for a given sample, the number of households who would purchase the new ketchup is different from the 5%.

It comes out to be a Bernulli distribution, however since the number of trials ($n=6000$) is suitable large, we may approximate it as a normal distribution.

• Determine the sample parameters

$$\bar x = \frac{335}{6000}=0.056, \; \; \; \; \sigma = \frac{\sigma_0}{\sqrt{6000}}=0.0089.$$

• Calculate the z-score statistic,

$$z= \frac{\bar x - \mu_0}{\sigma}=0.67.$$

• Check in the z table, the position of the critical value, (with a 2% significance level) for the hypothesis to be false.

• Compare the results

Hence, being the z-score outside the critical zone, we can assume the hypothesis to be true. Therefore the company should not conclude that there is an increased interest in the new product.

The questions:

First of all, is the presented solution correct? Then, is there another method to solve the problem? And, how could I improve the exposition of the solution?

It is misleading to say "we can assume the hypothesis is true." Even to say "we can assume the null hypothesis to be true," which is what you meant, is a bit misleading. Actually, what you can say is that "we cannot (based on this data) reject the null hypothesis in favor of $H_1$."
• (-1). I disagree that this approach is "almost perfect" - far from it. $\hat p_1=\frac{335}{6000}$ and $\hat p_2=0.05$ are both estimates. As such the test conducted should be a 2-sample proportions z-test. That is, $H_0 \, : \, p_1 \geq p_2$ and $H_1 \, : \, p_1 < p_2$, using $\hat p_1$ and $\hat p_2$. Commented May 26, 2017 at 3:00