Question

The following \(n=10\) observations are a sample from a normal population: $$ \begin{array}{llllllllll} 7.4 & 7.1 & 6.5 & 7.5 & 7.6 & 6.3 & 6.9 & 7.7 & 6.5 & 7.0 \end{array} $$ a. Find the mean and standard deviation of these data. b. Find a \(99 \%\) upper one-sided confidence bound for the population mean \(\mu\). c. Test \(H_{0}: \mu=7.5\) versus \(H_{\mathrm{a}}: \mu < 7.5 .\) Use \(\alpha=.01 .\) d. Do the results of part b support your conclusion in part c?

Short answer

Based on the given step-by-step analysis and solution, provide a short answer summarizing the results. In this exercise, we analyzed a sample of 10 numbers and calculated the mean and standard deviation of the sample as 7.05 and 0.623, respectively. The 99% one-sided upper confidence bound for the population mean was found to be 7.82. When we performed a hypothesis test, we failed to reject the null hypothesis (\(H_0: \mu = 7.5\)) at a significance level of 0.01. Since 7.5 is within the 99% confidence bound (7.82), this supports our conclusion that there is not enough evidence to reject the null hypothesis.

Step-by-step solution

a. Calculate the mean and standard deviation

First, we should calculate the mean \(\bar{X}\) and standard deviation \(s\) of the sample. The mean is calculated as follows: $$\bar{X} = \frac{\sum X_i}{n}$$ The standard deviation is calculated as follows: $$s = \sqrt{\frac{\sum(X_i - \bar{X})^2}{n-1}}$$ Using the given data, we get \(\bar{X} = \frac{7.4 + 7.1 + 6.5 + 7.5 + 7.6 + 6.3 + 6.9 + 7.7 + 6.5 + 7.0}{10} = 7.05\). Further calculating the standard deviation, we obtain \(s \approx 0.623\).

b. Find the upper bound of the 99% confidence interval

We will calculate the upper bound for the population mean (\(\mu\)) using the t-distribution. The upper bound formula is given by: $$\bar{X} + t_{\alpha, df} \cdot \frac{s}{\sqrt{n}}$$ where \(t_{\alpha, df}\) represents the critical value from the t-table at a given confidence level and degrees of freedom \(df=n-1\). For this problem, we have \(\alpha = 0.01\), and \(n = 10\), so \(df = 10 - 1 = 9\). Looking up the t-table (or using a calculator), we find \(t_{\alpha, df} = 2.821\). Plugging in the values, we get: $$7.05 + 2.821 \cdot \frac{0.623}{\sqrt{10}} \approx 7.82$$ Thus, the 99% one-sided upper confidence bound for the population mean is 7.82.

c. Perform the hypothesis test

We will perform a one-sample t-test for the given hypothesis with a significance level of \(\alpha = 0.01\). Our null hypothesis is \(H_0: \mu = 7.5\), and our alternative hypothesis is \(H_a: \mu < 7.5\). First, we calculate the t-statistic using the formula: $$t = \frac{\bar{X} - \mu_0}{\frac{s}{\sqrt{n}}}$$ Plugging in the values, we get: $$t = \frac{7.05 - 7.5}{\frac{0.623}{\sqrt{10}}} \approx -2.349$$ Now we will compare our t-statistic with the critical t-value. For a one-tailed test at \(\alpha = 0.01\) and \(df = 9\), the critical t-value is \(-2.821\). Since our t-statistic (-2.349) is greater than the critical t-value (-2.821), we fail to reject the null hypothesis.

d. Support for the conclusion in part c

In part c, we concluded that we couldn't reject the null hypothesis (\(H_0: \mu = 7.5\)) in favor of the alternative hypothesis (\(H_a: \mu < 7.5\)) at a significance level of \(\alpha = 0.01\). In part b, we found that the 99% one-sided upper confidence bound for the population mean was 7.82. Since 7.5 is within the confidence bound (7.82), it provides additional support for our conclusion in part c that we cannot reject the null hypothesis at a significance level of 0.01.

Key concepts

Confidence Interval

The confidence interval is a range of values that's used to estimate a population parameter. It's constructed so that, with a certain degree of confidence, the interval contains the true parameter. In this exercise, we are interested in finding the 99% upper one-sided confidence bound for the population mean.

What does this mean practically? Well, when we say a 99% confidence interval, we're saying if we were to take many repeated samples and compute their confidence intervals, then 99% of those intervals would contain the true mean. In this situation, calculating this confidence bound gives us an upper limit, suggesting the highest value that the population mean is likely to be.

• The formula used is:
  \[ \bar{X} + t_{\alpha, df} \cdot \frac{s}{\sqrt{n}} \]
• Here, \( \bar{X} \) is the sample mean, \( s \) the sample standard deviation, and \( t_{\alpha, df} \) the critical value from the t-distribution for the specific confidence level and degrees of freedom (\( df = n - 1 \)).
• This upper confidence bound helps in understanding the estimated upper limit beyond which the true mean is unlikely to fall.

Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions about a population based on sample data. It involves the comparison of two hypotheses: the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \). In this exercise, the null hypothesis states that the population mean \( \mu = 7.5 \) whereas the alternative hypothesis suggests that \( \mu < 7.5 \).

To perform this test, we calculate the t-statistic:
\[ t = \frac{\bar{X} - \mu_0}{\frac{s}{\sqrt{n}}} \]
Here, \( \mu_0 \) is the claimed mean value in the null hypothesis, and the rest are as defined before. The t-statistic tells us how far, in standard error units, our sample mean is from the claimed population mean.

• If the t-statistic falls in the critical region (determined by the significance level \( \alpha \)), we reject the null hypothesis.
• In our case, we observed that the calculated t-statistic (-2.349) was not extreme enough compared to the critical t-value (-2.821) to justify rejecting \( H_0 \).
• This means that, at the 0.01 significance level, we don't have enough evidence to conclude that \( \mu < 7.5 \).

Normal Distribution

Normal distribution is a fundamental concept in statistics, describing a symmetrical, bell-shaped distribution of data where most of the data points accumulate around the mean. Many statistical tests and methods, including those used in this exercise, rely on the assumption of normal distribution.

The reason this is important is because it allows us to utilize specific statistical tests and interpret results appropriately. The sample data in this exercise is assumed to come from a normally distributed population, which validates the use of the t-distribution in calculating the confidence interval and conducting the hypothesis test.

• When data is normally distributed, calculations such as the mean, standard deviation, and other statistics provide meaningful insights.
• Moreover, with a normal distribution, the central limit theorem implies that even with small sample sizes (like \( n=10 \) in this exercise), the sample mean will tend to be approximately normally distributed, thus facilitating inference.
• Understanding normal distribution is crucial as it helps in assessing the reliability of estimations and conclusions drawn from statistical analyses like the ones in the original exercise.