Question

According to experts, 10% of people are left-handed. a researcher claims that the figure is higher for people living in the town of cherryville. a random sample of 255 people living in cherryville yielded 39 who are left-handed. Test the researcher's claim at the 0.05 significance level.

Answer 1

To address the claim that Cherryville has a higher proportion of left-handed individuals than the general 10%, we calculate a one-sample z-test for proportions and compare the z-value to the critical value at a 0.05 significance level to potentially reject the null hypothesis.

Step-by-step explanation:

To test the researcher's claim that the proportion of left-handed people in Cherryville is higher than the generally accepted 10%, we can perform a one-sample z-test for proportions. The null hypothesis (H₀) states that the proportion of left-handed individuals in Cherryville is equal to 10% (the general population proportion), while the alternative hypothesis (H₁) suggests that this proportion is greater than 10%.

The test statistic for a one-sample z-test for proportions is calculated as follows:

z = (p_a - p_0) / √(p_0(1 - p_0) / n)

Where p_a is the sample proportion, p_0 is the hypothesized population proportion, and n is the sample size.

In this case, p_a = 9 / 255 and p_0 = 0.1. After calculating the z-value, we compare it to the critical value from the standard normal distribution for a 0.05 significance level. If the z-value is greater than the critical value, we reject the null hypothesis and accept the researcher's claim.