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The density of the earth relative to the density of water is known to be 5.517 g/cm \(^{3}\). Henry Cavendish, an English chemist and physicist \((1731-1810)\) was the first scientist to accurately measure the density of the earth. Following are 29 measurements taken by Cavendish in 1798 using a torsion balance. $$\begin{array}{lllllllllll}\hline 5.50 & 5.61 & 4.88 & 5.07 & 5.26 & 5.55 & 5.36 & 5.29 & 5.58 & 5.65 & 5.57 \\\5.53 & 5.62 & 5.29 & 5.44 & 5.34 & 5.79 & 5.10 & 5.27 & 5.39 & 5.42 & 5.47 \\\5.63 & 5.34 & 5.46 & 5.30 & 5.75 & 5.68 & 5.85 & & & & \\\\\hline\end{array}$$ a. What evidence do you have that the assumption of normality is reasonable? Explain. b. Is the mean of Cavendish's data significantly less than today's recognized standard? Use a 0.05 level of significance.

Short Answer

Expert verified
The evidence of data normality could come from visual inspection or statistical tests. The conclusion about the mean of Cavendish's data being significantly less than today's standard would depend on the results of the t-test.

Step by step solution

01

Normality Assumption

Firstly, the normality of Cavendish's measurements needs to be assessed. This can be done by creating a frequency distribution and observing the shape, using a 'Normal Probability Plot', or conducting a normality test such as the Shapiro-Wilk test. It is expected that all these tests should suggest the data is normally distributed.
02

Calculate the Mean of Cavendish's Data

This involves summing up the values of Cavendish's measurements and dividing by the number of measurements. This gives the average (mean) of Cavendish's Data.
03

Formulate Hypotheses for the T-test

The null hypothesis (H0) could be 'The mean of Cavendish's data is equal to the today's recognized standard (5.517 g/cm3)', and the alternative hypothesis (H1) 'The mean of Cavendish's data is less than the today's recognized standard'.
04

Perform the T-test

The t-test compares the mean of Cavendish's measurements with today's recognized standard to determine if there is a significant difference. A critical t-value can be obtained using the t-distribution table based on the chosen level of significance (0.05) and degrees of freedom (n-1). If the calculated t-value is less than the critical t-value, then the null hypothesis cannot be rejected at a 0.05 level of significance.

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Key Concepts

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

Normal Probability Plot
When analyzing the density measurements of the Earth taken by Henry Cavendish, it's crucial to determine whether the data follows a normal distribution. The Normal Probability Plot provides a graphical representation to assess this assumption of normality.

A normal probability plot is a scatter plot with the measurements on the y-axis and their expected z-scores on the x-axis, if the data were normally distributed. If the data points closely follow a straight line, this indicates that the data are normally distributed. This plot is especially useful because it allows for the visual identification of outliers or data skewness.

In Cavendish's case, examining a normal probability plot constructed using his 29 measurements would provide visual evidence regarding the normality of the data. Any significant deviations from a straight line might suggest non-normality which would then necessitate the use of non-parametric statistical methods.
Shapiro-Wilk Test
Complementing the visual assessment provided by the normal probability plot, the Shapiro-Wilk Test offers a more quantitative approach to testing normality. It is a widely used formal statistical test for assessing whether a set of data is normally distributed.

The Shapiro-Wilk test calculates a W statistic that represents how well the data conforms to a normal distribution. The lower the W statistic, the less likely the data is normally distributed. When applying this test to Cavendish's data, a significance level (alpha) is chosen, commonly 0.05. If the p-value from the test is less than alpha, the null hypothesis of normality is rejected, indicating the data may not be normally distributed.

For Cavendish's measurements, if the Shapiro-Wilk test yields a p-value greater than 0.05, this lends credence to the normality assumption, supporting the subsequent use of parametric tests such as the t-test.
T-test
Finally, to determine whether Cavendish's measurement of the Earth's density is significantly different from the recognized standard today, a T-test is utilized. The t-test is a statistical hypothesis test that compares the means of two samples or a sample mean to a known value, under the assumption that the underlying data distribution is normal.

In step 3 of the solution, we establish our null hypothesis (H0) that Cavendish's mean measurement is the same as today's standard, against the alternative hypothesis (H1) that it is less. Moving to step 4, the t-test calculation will compare his mean to the standard value, producing a t-statistic. If the calculated t-value is lower than the critical t-value from the t-distribution table, considering our degrees of freedom (n-1) and a 0.05 significance level, we do not reject H0. This means there's insufficient evidence to say that Cavendish's mean is significantly different from the standard.

The t-test is a powerful tool in this context, as it allows for conclusions about the reliability of historical measurements in light of current standards, assuming all test conditions are met.

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Most popular questions from this chapter

The full-time student body of a college is composed of \(50 \%\) males and \(50 \%\) females. Does a random sample of students \((30 \text { male, } 20\) female) from an introductory chemistry course show sufficient evidence to reject the hypothesis that the proportion of male and of female students who take this course is the same as that of the whole student body? Use \(\alpha=0.05\). a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

In the past the standard deviation of weights of certain 32.0 -oz packages filled by a machine was 0.25 oz. A random sample of 20 packages showed a standard deviation of 0.35 oz. Is the apparent increase in variability significant at the 0.10 level of significance? Assume package weight is normally distributed. a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

State the null hypothesis, \(H_{o}\), and the alternative hypothesis, \(H_{a},\) that would be used to test these claims: a. The standard deviation has increased from its previous value of 24. b. The standard deviation is no larger than 0.5 oz. c. The standard deviation is not equal to \(10 .\) d. The variance is no less than \(18 .\) e. The variance is different from the value of \(0.025,\) the value called for in the specs.

A winemaker has placed a large order for the no. 9 corks described in Applied Example 6.13 (p. 285 ) and is concerned about the number of corks that might have smaller diameters. During the corking process, the corks are squeezed down to 16 to \(17 \mathrm{mm}\) in diameter for insertion into bottles with an \(18 \mathrm{mm}\) opening. The cork then expands to make the seal. The winemaker wants the corks to be as tight as possible and is therefore concerned about any that might be undersized. The diameter of each cork is measured in several places, and an average diameter is reported for each cork. The cork manufacturer has assured the winemaker that each cork has an average diameter within the specs and that all average diameters have a normal distribution with a mean of \(24.0 \mathrm{mm}\). a. Why does it make sense for the diameter of the cork to be assigned the average of several different diameter measurements? A random sample of 18 corks is taken from the batch to be shipped and the diameters (in millimeters) obtained: $$\begin{array}{llllllll}\hline 23.93 & 23.91 & 23.82 & 24.02 & 23.93 & 24.17 & 23.93 & 23.84 & 24.13 \\\24.01 & 23.83 & 23.74 & 23.73 & 24.10 & 23.86 & 23.90 & 24.32 & 23.83 \\\\\hline\end{array}$$ b. The average diameter spec is "24 \(\mathrm{mm}+0.6 \mathrm{mm} /\) \(-0.4 \mathrm{mm} . "\) Does it appear this order meets the spec on an individual cork basis? Explain. c. Does the sample in part a show sufficient reason to doubt the truthfulness of the claim, that the mean average diameter is \(24.0 \mathrm{mm},\) at the 0.02 level of significance? A different sample of 18 corks was randomly selected and the diameters (in millimeters) obtained: $$\begin{array}{lllllllll}\hline 23.90 & 23.98 & 24.28 & 24.22 & 24.07 & 23.87 & 24.05 & 24.06 & 23.82 \\\24.03 & 23.87 & 24.08 & 23.98 & 24.21 & 24.08 & 24.06 & 23.87 & 23.95 \\\\\hline\end{array}$$ d. Does the preceding sample show sufficient reason to doubt the truthfulness of the claim, that the mean average diameter is \(24.0 \mathrm{mm},\) at the 0.02 level of significance? e. What effect did the two different sample means have on the calculated test statistic in parts c and d? Explain. f. What effect did the two different sample standard deviations have on the calculated test statistic in parts c and d? Explain.

The marketing research department of an instant-coffee company conducted a survey of married men to determine the proportion of married men who prefer their brand. Of the 100 men in the random sample, 20 prefer the company's brand. Use a \(95 \%\) confidence interval to estimate the proportion of all married men who prefer this company's brand of instant coffee. Interpret your answer.

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