Suppose a population is sampled and the sample mean is 16.8. The sample standard deviation is 6.0 I am to assume the sample is larger than 30) and the sample size is 65.
How do I calculate 95% and 98% confidence intervals.
State which interval wider and why please.
How would I also explain the results of the 95% confidence interval to my boss.
Please help
To calculate the confidence intervals, you can use the formula:
Confidence Interval = Sample Mean ± Critical Value * (Sample Standard Deviation / √Sample Size)
First, let's calculate the critical values for a 95% confidence level and a 98% confidence level. For a sample size greater than 30, you can use the Z-score to find the critical value. The Z-score corresponds to the desired confidence level.
For a 95% confidence level:
The critical value is approximately 1.96 (you can find this value from a Z-table or calculator).
For a 98% confidence level:
The critical value is approximately 2.33.
Now, let's calculate the confidence intervals:
95% Confidence Interval:
Sample Mean ± (1.96 * (Sample Standard Deviation / √Sample Size))
= 16.8 ± (1.96 * (6.0 / √65))
= 16.8 ± 1.96 * 0.74
= 16.8 ± 1.45
= (15.35, 18.25)
98% Confidence Interval:
Sample Mean ± (2.33 * (Sample Standard Deviation / √Sample Size))
= 16.8 ± (2.33 * (6.0 / √65))
= 16.8 ± 2.33 * 0.74
= 16.8 ± 1.72
= (15.08, 18.52)
The 98% confidence interval is wider than the 95% confidence interval. This is because a higher confidence level requires a larger critical value, which leads to a wider range of values in the confidence interval.
When explaining the results of the 95% confidence interval to your boss, you can say that with 95% confidence, the true population mean lies between 15.35 and 18.25. This means that if the sampling process was repeated many times and a new confidence interval was calculated each time, 95% of those intervals would contain the true population mean.