Final answer:
To test the marketing executive's claim, we set up the null and alternative hypotheses and use the P-value method. With a sample size of 1100 and an observed percentage of 74.8%, we calculate the test statistic z and find the P-value to be approximately 0.0219. Since the P-value is less than the significance level of 0.10, we reject the null hypothesis and conclude that the percentage of homes with Internet-connected TV devices is different from 78%.
Step-by-step explanation:
To test the marketing executive's claim, we can set up the null and alternative hypotheses. The null hypothesis (H0) assumes that the percentage of all homes with at least one Internet-connected TV device is equal to 78%. The alternative hypothesis (H1) assumes that the percentage is not equal to 78%. We will use the P-value method to test the claim.
Given that the sample size is 1100 and the observed percentage is 74.8%, we can calculate the test statistic z using the formula: z = (p - P) / sqrt(P * (1 - P) / n), where p is the observed percentage, P is the claimed percentage, and n is the sample size.
Substituting the given values, we have: z = (0.748 - 0.78) / sqrt(0.78 * (1 - 0.78) / 1100) = -2.296.
To find the P-value, we need to find the area under the standard normal curve to the left of -2.296. Using a standard normal table or a calculator, we find that the P-value is approximately 0.0219.
Since the P-value (0.0219) is less than the significance level (0.10), we reject the null hypothesis. This means that there is sufficient evidence to conclude that the percentage of all homes with at least one Internet-connected TV device is different from 78%.