Suppose that a sample of 100 independent draws from a normal distribution having unknown mean μ and known variance σ2 = 1 is observed. If the sample mean is 5, then the 95% confidence interval for μ is:
(4.804, 5.196)
Understanding Confidence Intervals for Normal Mean
This question asks us to calculate a 95% confidence interval for the unknown mean ($\mu$) of a normal distribution. We are given specific information about the distribution and a sample drawn from it.
- The population distribution is normal.
- The population variance ($\sigma^2$) is known and equals 1.
- The sample size ($n$) is 100.
- The sample mean ($\bar{x}$) is 5.
- The desired confidence level is 95%.
Since the population variance ($\sigma^2$) is known and the sample size is large ($n \ge 30$) (or the population is normal, which is given here), we use the z-distribution to construct the confidence interval for the population mean ($\mu$).
Formula for Confidence Interval (Known Variance)
The formula for a $(1 - \alpha) \times 100\%$ confidence interval for the population mean ($\mu$) when the population variance ($\sigma^2$) is known is:
\[ \bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}} \]Where:
- \(\bar{x}\) is the sample mean.
- \(\sigma\) is the population standard deviation (the square root of the variance).
- \(n\) is the sample size.
- \(z_{\alpha/2}\) is the z-score corresponding to the desired confidence level. It is the value such that the area to its right under the standard normal curve is \(\alpha/2\).
Step-by-Step Calculation
Let's plug in the given values into the formula.
- Identify the given values:
- Sample mean, \(\bar{x} = 5\).
- Population variance, \(\sigma^2 = 1\).
- Population standard deviation, \(\sigma = \sqrt{1} = 1\).
- Sample size, \(n = 100\).
- Confidence level = 95%, which means \(1 - \alpha = 0.95\).
- Calculate the significance level, \(\alpha\):
- \(\alpha = 1 - 0.95 = 0.05\).
- Find \(\alpha/2\):
- \(\alpha/2 = 0.05 / 2 = 0.025\).
- Find the critical z-score, \(z_{\alpha/2} = z_{0.025}\):
- For a 95% confidence interval, we need the z-score that leaves 0.025 area in the upper tail (and 0.025 in the lower tail), total \(\alpha = 0.05\) in both tails. This corresponds to a cumulative area of \(1 - \alpha/2 = 1 - 0.025 = 0.975\) to the left of \(z_{\alpha/2}\).
- Looking up 0.975 in the standard normal (z) table, we find that the corresponding z-score is 1.96. So, \(z_{0.025} = 1.96\).
- Calculate the standard error of the mean:
- Standard Error \( = \frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt{100}} = \frac{1}{10} = 0.1 \).
- Calculate the margin of error:
- Margin of Error \( = z_{\alpha/2} \times \text{Standard Error} = 1.96 \times 0.1 = 0.196 \).
- Construct the confidence interval:
- Lower bound \( = \bar{x} - \text{Margin of Error} = 5 - 0.196 = 4.804 \).
- Upper bound \( = \bar{x} + \text{Margin of Error} = 5 + 0.196 = 5.196 \).
The 95% confidence interval for \(\mu\) is \((4.804, 5.196)\).
Confidence Interval Result
Based on the sample data and the known variance, we are 95% confident that the true population mean (\(\mu\)) lies between 4.804 and 5.196.
Comparing with Options
Let's compare our calculated interval \((4.804, 5.196)\) with the given options:
| Option | Interval | Matches Calculation? |
|---|---|---|
| 1 | (4.84, 6.196) | No |
| 2 | (4.804, 5.196) | Yes |
| 3 | (4.23, 5.19) | No |
| 4 | (4.804, 6.196) | No |
Our calculated interval matches Option 2.
Revision Table: Key Concepts
| Concept | Description |
|---|---|
| Confidence Interval | A range of values, calculated from sample data, that is likely to contain an unknown population parameter (like the mean) with a certain level of confidence. |
| Confidence Level | The probability that a randomly selected confidence interval will contain the true population parameter. Commonly 90%, 95%, or 99%. |
| Significance Level (\(\alpha\)) | The probability of the confidence interval *not* containing the true population parameter. \(\alpha = 1 - \text{Confidence Level}\). |
| Margin of Error | The half-width of the confidence interval, calculated as \(z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\) (for known variance) or \(t_{\alpha/2, n-1} \frac{s}{\sqrt{n}}\) (for unknown variance). |
| Z-score (\(z_{\alpha/2}\)) | The critical value from the standard normal distribution used when the population variance is known or the sample size is large. |
Additional Information on Confidence Intervals
Constructing confidence intervals is a fundamental aspect of inferential statistics. It allows us to estimate population parameters based on sample data and quantify the uncertainty of our estimate.
- The width of the confidence interval is influenced by the confidence level, the standard deviation (or estimate), and the sample size.
- A higher confidence level (e.g., 99% vs. 95%) leads to a wider interval, as we need to be more certain that the interval contains the true parameter.
- A larger sample size ($n$) leads to a narrower interval, as the sample mean becomes a more precise estimate of the population mean.
- If the population variance ($\sigma^2$) is unknown and the sample size is small ($n < 30$), the t-distribution is used instead of the z-distribution.
- The interpretation of a 95% confidence interval is that if we were to take many random samples and compute a confidence interval for each, about 95% of those intervals would contain the true population mean.
Similar Questions
Which of the following is a merit of data tabulation?
In seasonal variations, the duration of time is not more than:
If a constant is added to each observation of a data set, then which of the following measures of dispersion will change?
Which of the following is NOT true for seasonal variation?
The analysis of variance technique was developed by:
For the series 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, the minimum possible value of \(\rm \Sigma_{i=1}^n(x_i-A)^2\) can be attained at:
Consider the following ANOVA table.
Source of variation Degrees of freedom The sum of Squares (SS) Mean SS F Ratio Treatments a b c 5 Error 12 d 20 Total 15 540
The values of a, b, c and d are, respectively:If the mean, mode and quartile deviation of a distribution is 2, 7 and 3, respectively, then Karl Pearson's coefficient of skewness is given by:
The component containing the overall upward or downward pattern of the data in an annual time series is:
If the 25th, 50th and 75th percentile of a frequency distribution are equal to 2, 3 and 4, respectively, then the distribution is:
Important Questions from Statistics
Methods of presentation of data are:
Consider 35 i.i.d. observations X1, X2, ...., X15 and Y1, Y2, ..., Y20. Let R be the Wilcoxon's rank sum statistic based on the ranks of the X' s in the combined sample. Then the expected value of R is
Let I, J > 5. Consider two-way ANOVA, where the observations satisfy the linear model
yij = α + βi + γj + εij, 1 ≤ i ≤ I, 1 ≤ j ≤ J
E(εij) = 0, Var(εij) = σ2, \(\sum_{i=1}^{!} β_i=\sum_{j=1}^j γ_j=0 .\)
In this set - up
Let X1,X2,...,Xn be a random sample from exponential distribution with mean θ. Which of the following is NOT a sufficient statistic for θ?
A newly developed algorithm for random number generation needs to be tested. The first step is to check whether the sequence of numbers generated can be considered a random sample from the uniform distribution on the interval (0, 1). Which of the following is an appropriate nonparametric test?
