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Chapter 10: Problem 5

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

Expert verified
Answer: Yes, the results support each other. The upper limit of the 99% confidence interval is 7.45, which is less than 7.5. Additionally, the hypothesis test rejected the null hypothesis, favoring the alternative hypothesis that the population mean is less than 7.5. Both results provide significant evidence to conclude that the population mean is less than 7.5.

Step by step solution

01

Calculate the mean and standard deviation of the sample

Using the given data points, we can calculate the mean by adding up all the values and then dividing by the number of values (10). Mean \(\bar{x} = \frac{7.4 + 7.1 + 6.5 + 7.5 + 7.6 + 6.3 + 6.9 + 7.7 + 6.5 + 7.0}{10} = 6.95\) Next, we will compute the standard deviation using the formula: \( s = \sqrt{\frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n-1}} \) Standard deviation \(s = \sqrt{\frac{(7.4-6.95)^2 + (7.1-6.95)^2 + (6.5-6.95)^2 + (7.5-6.95)^2 + (7.6-6.95)^2 + (6.3-6.95)^2 + (6.9-6.95)^2 + (7.7-6.95)^2 + (6.5-6.95)^2 + (7.0-6.95)^2}{9}} = \sqrt{\frac{2.83}{9}} = 0.56\) Now we have the mean and standard deviation of the sample: \(\bar{x} = 6.95\) and \(s = 0.56\).
02

Calculate the 99% upper one-sided confidence interval for the population mean

For this step, we will use the sample mean and standard deviation to calculate the 99% upper one-sided confidence interval for the population mean \(\mu\). Using t-distribution, our confidence interval can be calculated using the following formula: \(\bar{x} + t_{\alpha, df}\frac{s}{\sqrt{n}}\) First, we need to find the value \(t_{\alpha, df}\) from the t-distribution table, where \(\alpha = 0.01\) and degrees of freedom \(df = n - 1 = 10 - 1 = 9\). We find that \(t_{0.01,9} = 2.821\). Now we plug in the values into the formula: Confidence Interval \(= \bar{x} + t_{\alpha, df}\frac{s}{\sqrt{n}} = 6.95 + 2.821\frac{0.56}{\sqrt{10}} = 6.95 + 2.821 \cdot 0.177 = 7.45\) The 99% upper one-sided confidence interval for the population mean \(\mu\) is \(7.45\).
03

Perform hypothesis test for \(H_0: \mu = 7.5\) versus \(H_a: \mu < 7.5\) with \(\alpha = 0.01\)

We'll perform a one-sample t-test to test the hypothesis. First, calculate the test statistic using the formula: \(t = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}\), where \(\mu_0 = 7.5\). Test statistic \(t = \frac{6.95 - 7.5}{\frac{0.56}{\sqrt{10}}} = \frac{-0.55}{0.177} = -3.107\) Now, we need to find the critical value for a one-tailed test with \(\alpha = 0.01\) and degrees of freedom \(df = 9\). From the t-distribution table, we find that \(t_{crit} = -2.821\). Since our test statistic \(t = -3.107 < t_{crit} = -2.821\), we reject the null hypothesis \(H_0: \mu = 7.5\) in favor of the alternative hypothesis \(H_a: \mu < 7.5\).
04

Compare the results and conclusions of parts b and c

In part b, the 99% upper one-sided confidence interval for the population mean \(\mu\) is \(7.45\). In part c, we rejected the null hypothesis \(H_0: \mu = 7.5\) in favor of the alternative hypothesis \(H_a: \mu < 7.5\). The result of part b supports the conclusion of part c, since the upper confidence limit is less than the proposed mean value of \(7.5\). Both parts show that we have significant evidence to conclude that the population mean \(\mu\) is less than \(7.5\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Normal Distribution
A normal distribution, often called a bell curve, is a way to represent data. It looks like a hill, symmetrical around the center. Many real-world phenomena follow this pattern. In statistics, we use it to understand how data is spread out.

For example, if you measure many people's heights, most will be around the average with fewer people being a lot taller or shorter. This makes a normal distribution ideal for exploring our sample data.

Knowing that the data probably fits this pattern allows us to make predictions about the population it comes from. This can guide us when calculating metrics like the mean and standard deviation. It provides a basis for other important statistical concepts.
Confidence Interval Insights
A confidence interval is a range of values that estimates a population parameter, like the mean. It shows where the true parameter value might lie, given a certain level of confidence.

In our exercise, we calculated a 99% upper one-sided confidence interval. This tells us that we can be 99% confident the true mean is less than or equal to this calculated limit (7.45).

Why 99%? Higher confidence levels provide broader intervals because we want to be more sure our interval captures the true mean.
  • Upper one-sided interval: Only looks at values above the mean.
  • Calculated with the t-distribution for more accuracy in small samples.
This concept is crucial for understanding how confident we are in our results.
Basics of Hypothesis Testing
Hypothesis testing is a method used to decide if a statement about a population parameter is likely true. We set up a null hypothesis ( H_0 ) and an alternative hypothesis ( H_a ).

In our case, we tested if the mean ( H_0: 渭 = 7.5 gtrsim ) is equal to 7.5 versus it being less ( H_a: 渭 < 7.5 ). We used a one-sample t-test because we work with small samples and don鈥檛 know the population鈥檚 exact standard deviation.
  • Test statistic: Calculated from sample data to compare against critical values.
  • Significance level ( 伪 ): Probability of rejecting the null hypothesis when it's true. Here, 伪 = 0.01 means a 1% risk of error.
If the test statistic falls within a critical region, the null hypothesis is rejected. This gives us data-driven conclusions about our hypothesis.
Exploring the T-Distribution
The t-distribution is essential when working with small sample sizes. It adjusts for additional uncertainty. Unlike the normal distribution, its shape depends on sample size, specified by degrees of freedom ( df ).

In our exercise, we had df = n - 1 (10 - 1 = 9), meaning our data follows a t-distribution with 9 degrees of freedom.
  • More spread out than a normal distribution, especially with smaller samples.
  • As sample size increases, it gets closer to normal distribution.
We used it to find the critical value for our confidence interval and hypothesis test. This ensures our conclusions are applicable even with limited data, providing more reliable results.

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Most popular questions from this chapter

An experiment was conducted to compare the mean reaction times to two types of traffic signs: prohibitive (No Left Turn) and permissive (Left Turn Only). Ten drivers were included in the experiment. Each driver was presented with 40 traffic signs, 20 prohibitive and 20 permissive in random order. The mean time to reaction and the number of correct actions were recorded for each driver. The mean reaction times (in milliseconds) to the 20 prohibitive and 20 permissive traffic signs are shown here for each of the 10 drivers: \begin{tabular}{ll} Driver & Prohibitive & Permissive \\ \hline \end{tabular} \(\begin{array}{rrr}1 & 824 & 702 \\ 2 & 866 & 725 \\ 3 & 841 & 744 \\ 4 & 770 & 663 \\ 5 & 829 & 792 \\ 6 & 764 & 708 \\ 7 & 857 & 747 \\ 8 & 831 & 685 \\ 9 & 846 & 742 \\ 10 & 759 & 610\end{array}\) a. Explain why this is a paired-difference experiment and give reasons why the pairing should be useful in increasing information on the difference between the mean reaction times to prohibitive and permissive traffic signs. b. Do the data present sufficient evidence to indicate a difference in mean reaction times to prohibitive and permissive traffic signs? Use the \(p\) -value approach. c. Find a \(95 \%\) confidence interval for the difference in mean reaction times to prohibitive and permissive traffic signs.

The object of a general chemistry experiment is to determine the amount (in milliliters) of sodium hydroxide ( \(\mathrm{NaOH}\) ) solution needed to neutralize 1 gram of a specified acid. This will be an exact amount, but when the experiment is run in the laboratory, variation will occur as the result of experimental error. Three titrations are made using phenolphthalein as an indicator of the neutrality of the solution (pH equals 7 for a neutral solution). The three volumes of \(\mathrm{NaOH}\) required to attain a pH of 7 in each of the three titrations are as follows: \(82.10 .75 .75,\) and 75.44 milliliters. Use a \(99 \%\) confidence interval to estimate the mean number of milliliters required to neutralize 1 gram of the acid.

An advertisement for Albertsons, a supermarket chain in the western United States, claims that Albertsons has had consistently lower prices than four other fullservice supermarkets. As part of a survey conducted by an "independent market basket price-checking company," the average weekly total, based on the prices of approximately 95 items, is given for two different supermarket chains recorded during 4 consecutive weeks in a particular month. Week Albertsons Ralphs \(\begin{array}{lll}1 & 254.26 & 256.03 \\ 2 & 240.62 & 255.65 \\ 3 & 231.90 & 255.12\end{array}\) \(\begin{array}{lll}4 & 234.13 & 261.18\end{array}\) a. Is there a significant difference in the average prices for these two different supermarket chains? b. What is the approximate \(p\) -value for the test conducted in part a? c. Construct a \(99 \%\) confidence interval for the difference in the average prices for the two supermarket chains. Interpret this interval.

A paired-difference experiment was conducted to compare the means of two populations: Pairs Population 2 3 4 5 \(\begin{array}{llllll}1 & 1.3 & 1.6 & 1.1 & 1.4 & 1.7 \\\ 2 & 1.2 & 1.5 & 1.1 & 1.2 & 1.8\end{array}\) a. Do the data provide sufficient evidence to indicate that \(\mu_{1}\) differs from \(\mu_{2}\) ? Test using \(\alpha=.05\). b. Find the approximate \(p\) -value for the test and interpret its value. c. Find a \(95 \%\) confidence interval for \(\left(\mu_{1}-\mu_{2}\right) .\) Compare your interpretation of the confidence interval with your test results in part a. d. What assumptions must you make for your inferences to be valid?

The calcium (Ca) content of a powdered mineral substance was analyzed 10 times with the following percent compositions recorded: \(\begin{array}{lllll}.0271 & .0282 & .0279 & .0281 & .0268\end{array}\) \(\begin{array}{llll}.0271 & .0281 & .0269 & .0275 & .0276\end{array}\) a. Find a \(99 \%\) confidence interval for the true calcium content of this substance. b. What does the phrase " \(99 \%\) confident" mean? c. What assumptions must you make about the sampling procedure so that this confidence interval will be valid? What does this mean to the chemist who is performing the analysis?
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