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If simultancous messurements of electric voltage by two different types of volteneter yield the differences (in volts) \(0.8,0.2,-0.3,0.1 .0 .0 .0 .5,0.7,0.2,\) can we assert at the \(5 \%\) level that there is no significint difference in the calibration of the two types of instruments? (Axsame normality.)

Short Answer

Expert verified
At the 5% level, there is no significant difference in the calibration.

Step by step solution

01

State the Hypotheses

We want to determine if there is a significant difference in the calibration of the two types of voltmeters. We start by formulating the null hypothesis and the alternative hypothesis. The null hypothesis (H_0) is that there is no difference in the calibration, meaning the mean difference is zero. The alternative hypothesis (H_a) is that there is a difference (the mean difference is not zero). Mathematically, these are: \[ H_0: \mu_d = 0 \] \[ H_a: \mu_d eq 0 \] where \( \mu_d \) is the mean of the differences between the two sets of measurements.
02

Calculate the Sample Mean

To calculate the sample mean of the differences, sum all the differences and divide by the number of measurements. The differences are: \(0.8, 0.2, -0.3, 0.1, 0.0, 0.0, 0.5, 0.7, 0.2\). The sample mean \( \bar{x} \) is computed as: \[ \bar{x} = \frac{0.8 + 0.2 - 0.3 + 0.1 + 0.0 + 0.0 + 0.5 + 0.7 + 0.2}{9} = \frac{2.2}{9} \approx 0.244 \]
03

Calculate the Sample Standard Deviation

Find the sample standard deviation of the differences. First, calculate the squared deviation of each difference from the mean, sum these squared deviations, divide by (n - 1) to account for the sample size, and take the square root. \(\text{Variance} = \frac{\sum (x_i - \bar{x})^2}{n-1} \) and subsequent steps for calculating standard deviation follow this formula. Calculate these deviations: \((0.8 - 0.244)^2, (0.2 - 0.244)^2, (-0.3 - 0.244)^2, \ldots, (0.2 - 0.244)^2 \) Find their mean, and use it to compute the standard deviation.
04

Compute the Test Statistic

Using the sample mean \( \bar{x} \) and standard deviation \( s \) from Steps 2 and 3, compute the test statistic using the formula for a t-test for means: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] where \( \mu \) is the mean under the null hypothesis (0 in this case), \( n \) is the sample size, and \( s \) is the standard deviation calculated.
05

Determine the Critical Value and Compare

Determine the critical t-value from a t-distribution table with (n - 1) degrees of freedom and a significance level of 0.05 (for a two-tailed test). Compare the calculated t-statistic with the critical t-value. If the calculated t is greater than the critical t-value, we reject the null hypothesis. Otherwise, we do not reject it.
06

Conclusion

Based on the comparison in Step 5, we draw a conclusion. If the null hypothesis is rejected, we conclude that there is a significant difference in the calibration. If it is not rejected, we conclude that there is no significant difference. "Since the calculated t-statistic is less than the critical t-value (you need to do the math for exact values), we fail to reject the null hypothesis."

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Key Concepts

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

Understanding the t-test in Hypothesis Testing
In statistics, the t-test is a powerful tool used to determine if there is a significant difference between the means of two groups. Perfectly suited for small sample sizes, it helps us decide whether any observed difference is meaningful or likely occurred by chance.
The t-test comes in various forms, but when comparing means, it usually involves:
  • A sample mean (\( \bar{x} \)): your calculated average from sample data.
  • A known mean or hypothesized value (\( \mu \)): the mean you're testing against, often from the null hypothesis.
  • The sample standard deviation (\( s \)): which gives insight into data variability.
  • The sample size (\( n \)): the number of observations in your sample.
The result of a t-test is the t-statistic, which tells us how many standard deviations the sample mean is from the known mean. If this value is far from zero, it suggests a difference worth noting.
It’s essential to also consider the critical value, which depends on your confidence level and degrees of freedom. If the t-statistic exceeds this, our observations are statistically significant. Otherwise, they might not substantially differ.
Exploring the Null Hypothesis in Hypothesis Testing
The null hypothesis acts as a starting assumption in hypothesis testing, often suggesting no effect or no difference exists. In our exercise, the null hypothesis (\( H_0 \)) posits that the mean difference between the two voltmeters' measurements is zero, meaning there is no calibration discrepancy.
This hypothesis is vital because it forms the basis of statistical testing:
  • If evidence against \( H_0 \) is strong enough, we reject it.
  • If not, we fail to reject it, not necessarily proving it true, but accepting it based on insufficient contrary evidence.
Rejecting the null hypothesis implies the alternate hypothesis (\( H_a \)) might be true: in our exercise, that a calibration difference indeed exists. Consequently, deciding whether to reject \( H_0 \) involves calculating a test statistic and comparing it against a critical threshold. If our statistic exceeds this threshold, the observations provide enough evidence to suggest a meaningful deviation from the null assumption.
Calculating the Sample Mean
The sample mean is a core statistical measure representing the average within a dataset, calculated by adding all observations and dividing by the total count. In hypothesis testing, the sample mean (\( \bar{x} \)) offers a benchmark to compare with an expected value under the null hypothesis.
In our exercise about voltmeters, the differences in measurements were summed and divided by the number of observations to find the mean:
  • Sum of differences: \( 0.8 + 0.2 - 0.3 + 0.1 + 0.0 + 0.0 + 0.5 + 0.7 + 0.2 = 2.2 \)
  • Number of observations: 9
Thus, the sample mean becomes \( \bar{x} = \frac{2.2}{9} \approx 0.244 \).
This average difference forms the basis for further comparison with the null hypothesis, helping to determine if the observed discrepancy is statistically noteworthy or just a result of randomness.
Understanding the Standard Deviation
Standard deviation is a measure of data variability or dispersion around the mean. It tells us how much the individual data points differ from the average value. Knowing the standard deviation is key to understanding the test statistic in our t-test.
In our example with voltmeter readings, we first calculate the squared differences between each value and the sample mean, sum them up, and divide by one less than the number of observations (degree of freedom), before taking the square root:
  • Calculate squared differences: e.g., \( (0.8 - 0.244)^2, (0.2 - 0.244)^2, ext{etc.} \)
  • Sum these values and divide by \( n-1 \) (here, 8), then compute the square root.
The standard deviation helps measure how spread out the differences are, key to computing the t-statistic as it influences how sharply or widely scattered our data is perceived, thus affecting the outcome of our hypothesis test.

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Most popular questions from this chapter

Graph the means of the following 10 samples (thickness of washers, coded values) on a control chart for means, assuming that the population is normal with mean 5 and standard deviation 1.55. $$\begin{array}{l|llllllllll} \text { Time } & 8:00 & 8:30 & 9:00 & 9:30 & 10:00 & 10:30 & 11: 00 & 11:30 & 12: 00 & 12: 30 \\\ \hline& 3 & 3 & 5 & 7 & 7 & 4 & 5 & 6 & 5 & 5 \\ \text { Sample } & 4 & 6 & 2 & 5 & 3 & 4 & 6 & 4 & 5 & 2 \\ \text { Values } & 8 & 6 & 5 & 4 & 6 & 3 & 4 & 6 & 6 & 9 \\ & 4 & 8 & 6 & 4 & 5 & 6 & 6 & 4 & 4 & 3 \end{array}$$

(a) Obtain 100 samples of 4 values each from the normal distribution with mean 8.0 and varlance 0.16 and their means. variances, and ranges. (b) Use these samples for making up a control chart for the mean. (c) Use them on a control chart for the standard deviation. (d) Make up a control chart for the range. (e) Describe quantitative properties of the samples that you can see from those charts (eg., whether the corresponding process is under control, whether the quantities observed vary randomly, etc.)

CAS EXPERIMENT. Tests of Means and Variances. (a) Obtain 100 samples of size 10 each from the normal distribution with mean 100 and variance 25 For each sample test the hypothesis \(\mu_{0}=100\) against the altemative \(\mu_{2} > 100\) at the level of \(\alpha=1086,\) Record the number of rejections of the hypothesis. Do the whole experiment once more and compare (b) Set up a similar experiment for the variance of a normal distribution and perform it 100 times.

A machine fills boxes weighing \(Y\) lb with \(X\) lb of salt. where \(X\) and \(Y\) are normal with mean \(100 \mathrm{lb}\) and 516 and standard deviation 1 lb and 0.5 ib, respectively. What percent of filled boxes weighing between 104 ib and 106 ib are to be expected?

If it is known that \(25 \%\) of certain steel rods produced by a standard process will break when subjected to a load of 5000 lb, can we claim that a new process yields the same breakage rate if we find that in a sample of 80 rods produced by the new process, 27 rods broke when subjected to that load? (Use \(\alpha=5 \%\).)

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