Question

Past records from a large supermarket show that 20% of people who buy chocolate bars buy the family size bars. On one particular day a random sample of 30 people was taken from those that had bought chocolate bars and 2 of them were found to have bought a family size bar.
1) test at the 5 % significance level, whether or not the proportion of people who bought a family size bar of chocolate that day had decreased. State your hypothesis clearly.

Answer 1

Answer on Question #79712 – Math – Statistics and Probability

Question

Past records from a large supermarket show that 20% of people who buy chocolate bars buy the family size bars. On one particular day a random sample of 30 people was taken from those that had bought chocolate bars and 2 of them were found to have bought a family size bar.

1) test at the 5% significance level, whether or not the proportion of people who bought a family size bar of chocolate that day had decreased. State your hypothesis clearly.

Solution

The following information is provided: The sample size is $N = 30$, the number of favorable cases is $X = 2$, and the sample proportion is $\bar{p} = \frac{X}{N} = \frac{2}{30} = 0.0667$, and the significance level is $\alpha = 0.05$

The following null and alternative hypotheses need to be tested:

$$H_0: p = 0.2$$ $$H_\alpha: p < 0.2$$

This corresponds to a left-tailed test, for which a z-test for one population proportion needs to be used.

Based on the information provided, the significance level is $\alpha = 0.05$, and the critical value for a left-tailed test is $z_c = -1.64$.

The rejection region for this left-tailed test is $R = \{z: z < -1.64\}$

The z-statistic is computed as follows:

$$z = \frac{\bar{p} - p_0}{\sqrt{p_0(1 - p_0)/n}} = -1.826$$

Since it is observed that $z = -1.826 < z_c = -1.64$, then one concludes that the null hypothesis is rejected.

Using the P-value approach: The p-value is 0.0339, and since $0.0339 < 0.05$, one concludes that the null hypothesis is rejected.

Conclusion

The null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion $p$ is less than $p_0 = 0.2$ at the $\alpha = 0.05$ significance level.

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