Please help- A random sample of n=500 households where 150 have combined cable,phone, and internet service. Firrst, calculate p-hat, the sample proportion of households having such service. Then use a confidence level for pi, the proportion of all households in this area that have this service: use a confidence level of 90%
(-0.0371, 0.3715)
To calculate the sample proportion, p-hat, you need to divide the number of households with combined cable, phone, and internet service (150) by the total number of households in the sample (500).
p-hat = Number of households with the service / Total number of households in the sample
= 150 / 500
= 0.3
So, the sample proportion (p-hat) is 0.3.
To determine the confidence interval for the population proportion, pi, you can use the formula:
Confidence Interval = p-hat ± z * sqrt((p-hat * (1- p-hat)) / n)
Here, z represents the critical value for the desired confidence level. For a 90% confidence level, the critical value can be found using statistical tables or calculated using software. In this case, assuming a normal distribution and a two-tailed test, the z-value for a 90% confidence level is approximately 1.645.
Now, let's plug the values into the formula:
Confidence Interval = 0.3 ± 1.645 * sqrt((0.3 * (1-0.3)) / 500)
Calculating the square root and simplifying the equation:
Confidence Interval = 0.3 ± 1.645 * sqrt(0.21 / 500)
Confidence Interval = 0.3 ± 1.645 * 0.0204
Confidence Interval = 0.3 ± 0.0336
Therefore, the confidence interval for the population proportion, pi, is (0.2664, 0.3336) at a 90% confidence level. This means we can be 90% confident that the true proportion of households with combined cable, phone, and internet service in the entire area falls within this interval.