a fizzy water manufacturer advertises that its bottles hold 12 oz. of liquid. They want to estimate how many of their bottle hold less than that amount. They randomly select 60 filled bottles off the assembling line and find that 4 of them hold less than 12 oz. of liquid. Find the 5% confidence level margin of error for the sample proportion. Use x=1.960. Round your answer to the nearest tenth of a percent.
To find the 5% confidence level margin of error for the sample proportion, we need to calculate the standard error of the sample proportion first.
The formula for the standard error of the sample proportion is given by:
SE = sqrt((p * (1 - p)) / n),
where p is the sample proportion and n is the sample size.
In this case, the sample size is 60 bottles, and 4 out of the 60 bottles hold less than 12 oz. So we can calculate the sample proportion as:
p = 4/60 = 0.0667.
Substituting these values into the formula, we get:
SE = sqrt((0.0667 * (1 - 0.0667)) / 60) = 0.0580 (rounded to four decimal places).
Next, we need to calculate the critical value for a 5% confidence level. The critical value corresponds to a z-score, and at a 5% confidence level, the z-score is 1.960.
Finally, we calculate the margin of error using the formula:
ME = x * SE,
where x is the critical value and SE is the standard error.
Substituting the values, we have:
ME = 1.960 * 0.0580 = 0.1137 (rounded to four decimal places).
The margin of error for the sample proportion is 0.1137, which means that at a 5% confidence level, the true proportion of bottles holding less than 12 oz could be 0.1137 higher or lower than the observed sample proportion of 0.0667.
Rounded to the nearest tenth of a percent, the margin of error is 11.4%.