There are 250 computer programmers employed at Computers.com, Inc. A sample of 50 programmers revealed that 30 graduated with a four-year college degree. Construct the 95 percent confidence interval for the proportion of all programmers who graduated from a four-year university.
Well, it seems like Computers.com, Inc. has their own version of "The College Code". Anyway, let's compute that confidence interval for you.
First, we need to calculate the sample proportion. So, out of the 50 sampled programmers, 30 graduated from a four-year college. That means the sample proportion is 30/50 = 0.6.
Next, we'll determine the margin of error. Since we're aiming for a 95% confidence interval, we have to consider the critical value, which is approximately 1.96 for large sample sizes.
The margin of error is then given by: Margin of Error = Critical Value * Standard Error
Standard Error = sqrt[(Sample Proportion * (1 - Sample Proportion)) / Sample Size]
Plugging in the values, we have:
Standard Error = sqrt[(0.6 * 0.4) / 50] ≈ 0.085
Now, let's calculate the margin of error:
Margin of Error = 1.96 * 0.085 ≈ 0.1666
To construct the confidence interval, we'll use the formula: Confidence Interval = Sample Proportion ± Margin of Error
Confidence Interval = 0.6 ± 0.1666.
Therefore, the 95% confidence interval is approximately (0.4334, 0.7666).
So, based on this sample, we can be 95% confident that the proportion of all programmers at Computers.com, Inc. who graduated from a four-year college is somewhere between 43.34% and 76.66%. Keep coding, and may your programs be bug-free and full of humor, just like this bot!
To construct a confidence interval for the proportion of all programmers who graduated from a four-year university, we can use the formula:
Confidence interval = sample proportion ± critical value × standard error
First, let's calculate the sample proportion:
Sample proportion = number of programmers with a four-year college degree / sample size
= 30 / 50
= 0.6
Next, let's calculate the standard error:
Standard error = √[ (sample proportion × (1 - sample proportion)) / sample size ]
= √[ (0.6 × (1 - 0.6)) / 50 ]
= √[ (0.6 × 0.4) / 50 ]
= √(0.024 / 50)
≈ √(0.00048)
≈ 0.022
The critical value corresponds to the desired level of confidence. For a 95% confidence interval, the critical value can be determined using a standard normal distribution or a z-table. For this example, the critical value for a 95% confidence interval is approximately 1.96.
Now we can construct the confidence interval:
Confidence interval = 0.6 ± 1.96 × 0.022
Calculating the lower limit:
Lower limit = 0.6 - 1.96 × 0.022
≈ 0.6 - 0.04312
≈ 0.55688
Calculating the upper limit:
Upper limit = 0.6 + 1.96 × 0.022
≈ 0.6 + 0.04312
≈ 0.64312
Therefore, the 95 percent confidence interval for the proportion of all programmers who graduated from a four-year university is approximately 0.5569 to 0.6431.
To construct a confidence interval for a proportion, we can use the formula:
CI = p̂ ± Z * sqrt((p̂ * (1 - p̂)) / n)
where:
CI is the confidence interval,
p̂ is the sample proportion,
Z is the z-score corresponding to the desired confidence level, and
n is the sample size.
In this case, we have:
p̂ = 30/50 = 0.6 (proportion of programmers who graduated from a four-year university)
n = 50 (sample size)
To find the z-score corresponding to a 95 percent confidence level, we can look it up in a standard normal distribution table or use a calculator. The z-score for a 95 percent confidence level is approximately 1.96.
Now we can substitute the values into the formula:
CI = 0.6 ± 1.96 * sqrt((0.6 * (1 - 0.6)) / 50)
Calculating this expression gives:
CI ≈ 0.6 ± 1.96 * sqrt(0.24 / 50)
CI ≈ 0.6 ± 1.96 * sqrt(0.0048)
CI ≈ 0.6 ± 1.96 * 0.0693
CI ≈ 0.6 ± 0.1355
Therefore, the 95 percent confidence interval for the proportion of all programmers who graduated from a four-year university is approximately 0.4645 to 0.7355.