# 8.1.2.1.3 - Example: Ice Cream

8.1.2.1.3 - Example: Ice Cream

Research Question: Is the percentage of Creamery customers who prefer chocolate ice cream over vanilla less than 80%?

In a sample of 50 customers 60% preferred chocolate over vanilla.

1. Check assumptions and write hypotheses

$np_0 = 50(0.80) = 40$

$n(1-p_0)=50(1-0.80) = 10$

Both $np_0$ and $n(1-p_0)$ are at least 10. We can use the normal approximation method.

This is a left-tailed test because we want to know if the proportion is less than 0.80.

$H_{0}\colon p=0.80$
$H_{a}\colon p<0.80$

2. Calculate the test statistic
Test statistic: One Group Proportion

$z=\dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}$

$\widehat{p}$ = sample proportion
$p_{0}$ = hypothesize population proportion
$n$ = sample size

$\widehat{p}=0.60$, $p_{0}=0.80$, $n=50$

$z= \dfrac{\widehat{p}- p_0 }{\sqrt{\frac{p_0 (1- p_0)}{n}}}= \dfrac{0.60-0.80}{\sqrt{\frac{0.80 (1-0.80)}{50}}}=-3.536$

Our $z$ test statistic is -3.536.

3. Determine the p-value

This is a left-tailed test so we need to find the area to the right of our test statistic, $z=-3.536$.

From the Minitab Express output above, the p-value is 0.0002031

4. Make a decision

$p \leq.05$, therefore our decision is to reject the null hypothesis.

5. State a "real world" conclusion

Yes, there is evidence that the percentage of all Creamery customers who prefer chocolate ice cream over vanilla is less than 80%.

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