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Suppose we have a sample with \(\mathrm{n}=10\) and \(\underline{\mathrm{Y}}=50 . \mathrm{We}\) wish to test \(\mathrm{H}_{0}: \mu=47\) against \(\mathrm{H}_{1}: \mu \neq 47 .\) We would like to know the probability given \(\mu=47\) of observing a random sample from the population with \(\underline{Y}=50 .\) We will reject \(\mathrm{H}_{0}\) if the probability is less than \(\alpha=.05\) that \(\mathrm{H}_{0}\) is true when \(\underline{Y}=50\). Assume for this sample that \(\sum(\mathrm{Y}-\mathrm{Y})^{2}=99.2250 .\) Calculate the estimated population variance, the estimated \(\sigma^{2} \underline{\mathrm{Y}}\) and the estimated slandered error of the mean. Then calculate the t statistic and determine whether \(\mathrm{H}_{0}\) can be rejected. Also, suppose \(\alpha=.01\). Within what limits may \(\underline{Y}\) vary without our having to reject \(\mathrm{H}_{0}\) ?

Step by step solution

01

Calculate the sample variance

The sample variance s虏 is calculated using the formula: \( s^2 = \frac{\sum(Y-Y虅)^2}{n-1} \) Where n is the sample size, Y are the observations, and Y虅 is the sample mean. In this case, n=10, and 危(Y-Y虅)虏=99.2250. \( s^2 = \frac{99.2250}{10-1} = \frac{99.2250}{9} = 11.0250 \)

02

Calculate the estimated variance of Y虅

The estimated variance of Y虅, which is denoted as 蟽虏Y虅, can be found using the formula: \( \sigma_{Y虅}^2 = \frac{s^2}{n} \) In this case, s虏=11.0250, and n=10. \( \sigma_{Y虅}^2 = \frac{11.0250}{10} = 1.1025 \)

03

Calculate the estimated standard error of the mean

The estimated standard error of the mean (饾憼饾憣虅) is the square root of the estimated variance of Y虅: \( s_{Y虅} = \sqrt{\sigma_{Y虅}^2} = \sqrt{1.1025} = 1.05 \)

04

Calculate the t-statistic

The t-statistic can be calculated using the formula: \( t = \frac{Y虅 - \mu}{s_{Y虅}/\sqrt{n}} \) In this case, Y虅=50, 碌=47, sY虅=1.05, and n=10. \( t = \frac{50 - 47}{1.05/\sqrt{10}} = \frac{3}{1.05/3.16} = 2.857 \)

05

Determine whether to reject H鈧�

Using a t-distribution table or calculator, we can find the critical t-value for a two-tailed test with 伪=.05 and degrees of freedom (df) equal to n-1 (10-1=9). In this case, the critical t-value is approximately 2.262. Since our calculated t-statistic (2.857) is greater than the critical t-value (2.262), we reject H鈧� at 伪=.05. For 伪=.01, the critical t-value is approximately 3.250. Since our calculated t-statistic (2.857) is less than this critical t-value, we fail to reject H鈧� at 伪=.01.

06

Calculate the limits for Y虅 at 伪=.01

To find the limits for Y虅 within which we do not reject H鈧� at 伪=.01, we use the formula: Y虅 = 碌 卤 (t * sY虅 / 鈭歯) Where t is the critical t-value for 伪=.01 and degrees of freedom 9, which is approximately 3.250. Lower Limit: Y虅 = 47 + (-3.250 * 1.05 / 鈭�10) = 47 + (-1.086) = 45.914. Upper Limit: Y虅 = 47 + (3.250 * 1.05 / 鈭�10) = 47 + (1.086) = 48.086. Therefore, if Y虅 varies between 45.914 and 48.086, we will not reject H鈧� at 伪=.01.

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Key Concepts

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

t-statistic

When conducting a hypothesis test, one crucial element is calculating the t-statistic. This value helps determine how far the sample mean is from the hypothesized population mean in relation to the standard error. The formula for the t-statistic is:\[ t = \frac{\bar{Y} - \mu}{s_{Y虅}/\sqrt{n}} \]Where:

• \(\bar{Y}\) is the sample mean.
• \(\mu\) is the population mean hypothesized in \(H_0\).
• \(s_{Y虅}\) is the estimated standard error of the mean.
• \(n\) is the sample size.

To reject the null hypothesis \(H_0\), the t-statistic should be greater than the critical t-value for a specified significance level \(\alpha\). If the calculated t-statistic of 2.857 exceeds the critical t-value of 2.262 at \(\alpha = 0.05\), \(H_0\) is rejected, suggesting the sample provides enough evidence to support the alternative hypothesis. However, at a stricter \(\alpha = 0.01\), the critical t-value increases to 3.250, where \(H_0\) is not rejected.

sample variance

The sample variance is key to understanding how much variance or spread exists among sample data points. It provides insight into the reliability of the sample mean as an estimate of the population mean. For the given sample:\[ s^2 = \frac{\sum (Y - \bar{Y})^2}{n-1}\]With an example sample size \(n = 10\) and \(\sum (Y-\bar{Y})^2 = 99.2250\), compute:\[ s^2 = \frac{99.2250}{9} = 11.0250\]Here, \(n-1\) is used instead of \(n\) because it provides an unbiased estimate of the population variance from a sample. This is known as Bessel's correction.

standard error of the mean

The standard error of the mean (SEM) indicates how much the sample mean \(\bar{Y}\) is expected to vary from the true population mean \(\mu\). It is derived from the sample variance \(s^2\) and calculated as follows:First, calculate the estimated variance of the sample mean:\[ \sigma_{\bar{Y}}^2 = \frac{s^2}{n}\]Here, substituting \(s^2 = 11.0250\) and \(n = 10\), you get:\[ \sigma_{\bar{Y}}^2 = \frac{11.0250}{10} = 1.1025\]Then, the standard error is the square root of this variance:\[ s_{\bar{Y}} = \sqrt{1.1025} \approx 1.05\]A smaller SEM indicates that the sample mean is a more accurate reflection of the population mean. SEM plays a critical role in forming confidence intervals and conducting hypothesis tests, which help infer about the population from your sample.