Refer to Exercise 6.44 (p. 356), in which 50 consumers taste-tested a new snack food. Their responses (where 0 = do not like; 1 = like; 2 = indifferent) are reproduced below

1. Test \({H_0}:p = .5\) against \({H_0}:p > .5\), where p is the proportion of customers who do not like the snack food. Use \(\alpha = 0.10\).
   1 0 0 1 2 0 1 1 0 0 0 1 0 2 0 2 2 0 0 1 1 0 0 0 0 1 0 2 0 0 0 1 0 0 1 0 0 1 0 1 0 2 0 0 1 1 0 0 0 1

The number of sample size is 50.

The hypothesis are given by

\(\begin{aligned}{H_0}:p = .5\\{H_a}:p > .5\end{aligned}\)

A null hypothesis is a statistical supposition that claims there is no difference between specific features of a population as well as data-generating activity. The alternate hypothesis asserts that there is a distinction. Hypothesis test enables you to reject a null hypothesis with a particular confidence level.

The proportion of customers who do not like the snack food is given below

\(\begin{aligned}\hat p &= \frac{{29}}{{50}}\\ &= 0.58\end{aligned}\)

The test statistic is calculated as

\(\begin{aligned}z &= \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\\ &= \frac{{0.58 - 0.5}}{{\sqrt {\frac{{.5 \times .5}}{{50}}} }}\\ &= \frac{{0.08}}{{0.07}}\\ &= 1.142\end{aligned}\)

Therefore, the test statistic is 1.142.

Rejection region \(z > {z_{0.1}} = 1.282\)

Therefore, the calculated z-value does not fall in the rejection region. We do not reject the null hypothesis.