Chocolate and Happiness Use the results from part (b) of Cumulative Review Exercise 2 to test the claim that when asked, more than 80% of women say that chocolate makes them happier. Use a 0.01 significance level.

It is given that a survey was sponsored by a chocolate company.

Out of 1708 women who were surveyed, 85% of them said that chocolate made them happier.

The null hypothesis for testing the claim is as follows:

The population percentage of women who say that chocolate makes them happier is equal to 80%.

\({H_0}:p = 0.80\)

The alternative hypothesis for testing the claim is as follows:

The population percentage of women who say that chocolate makes them happier is more than 80%.

\({H_0}:p > 0.80\)

The test is right-tailed.

Let\(\hat p\)denote the sample proportion ofwomen who said that chocolate makes them happier.

Here,

\(\begin{aligned}{c}\hat p = 85\% \;\\ = 0.85\end{aligned}\)

Here, p=0.80.

Thus,

\[\begin{aligned}{c}q = 1 - p\\ = 1 - 0.80\\ = 0.20\end{aligned}\]

Since the sample size (n) equal to 1708 is large, the value of the z-score is computed as follows:

\[\begin{aligned}{c}z = \frac{{\hat p - p}}{{\sqrt {\frac{{pq}}{n}} }}\;\;\; \sim N\left( {0,1} \right)\\ = \frac{{0.85 - 0.80}}{{\sqrt {\frac{{\left( {0.80} \right)\left( {0.20} \right)}}{{1708}}} }}\\ = 5.166\end{aligned}\]

Thus, the test statistics, z is 5.166.

Referring to standard normal table,

The critical value of z at 0.01 level of significance for a right-trailed test is equal to 2.3263.

The corresponding p-value is equal to 0.000.

Since the absolute value of the z-score is greater than the critical value and the p-value is less than 0.01, the null hypothesis is rejected.

There is enough evidence to conclude that the percentage of women who say that chocolate makes them happier is more than 80%.