The new Twinkle bulb is being developed to last more than 1000 hours. A random sample of 100 of these new bulbs is selected from the production line. It was found that 48 lasted more than 1000 hours. Find the point estimate for the population proportion, the margin of error for a 95% confidence and then construct to the 95% confidence interval for the population proportion p.
Question: Find the margin of error E.
Round E to three decimal places
Answer:
E = zc �ãpq/n
= 1.96 �ã(0.48 x 0.52)/100
= 0.098
Question: Construct a 95% confidence interval for the population proportion p of all Twinkle bulbs.
Answer:
0.48 - 0.098 < p < 0.48 + 0.098
To find the margin of error (E), you can use the formula:
E = zc * sqrt((p * q) / n)
Where:
- zc is the critical value for the desired confidence level (95% confidence level corresponds to a z-score of approximately 1.96)
- p is the point estimate for the population proportion (48/100 = 0.48)
- q is the complement of p (1 - 0.48 = 0.52)
- n is the sample size (100)
Plugging in the values, we get:
E = 1.96 * sqrt((0.48 * 0.52) / 100)
E ≈ 0.098
So, the margin of error (E) is approximately 0.098.
To construct a 95% confidence interval for the population proportion (p), you can use the formula:
p - E < p < p + E
Replacing the values:
0.48 - 0.098 < p < 0.48 + 0.098
Simplifying:
0.382 < p < 0.578
Therefore, the 95% confidence interval for the population proportion (p) of all Twinkle bulbs is approximately 0.382 to 0.578.