The data of the problem is provided as follows:
$$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text { Production-line workers }\left(X_{1}\right) & 4 & 0 & 6 & 8 & 3 & 11 & 13 & 5 \\ \hline \text { Office workers }\left(X_{2}\right) & 9 & 2 & 7 & 1 & 4 & 7 & 9 & 8 \\ \hline \end{array} $$
Here, equal variance of t-test is used. Firstly, set the hypothesis as shown below:
State the null hypothesis:
\(H_{0}\) : There is no difference in days absent between the two groups of workers.
Against the alternative hypothesis:
\(H_{1}:\) There is significant difference in days absent between the two groups of workers.
The test statistic used is:
\(t=\frac{\left(\bar{x}_{1}-\bar{x}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{\sqrt{s_{p}^{2}\left(\frac{1}{n_{1}}+\frac{1}{n_{2}}\right)}}\)
where \(s_{p}^{2}=\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}\)
and \(n_{1}\left(\right.\) or \(\left.n_{2}\right)=8\)
The mean of samples is:
\(\begin{aligned} \bar{X}_{1} &=\frac{\sum_{i=1}^{8} X_{i}}{n_{1}} \\ &=\frac{4+0+6 \ldots .+5}{8} \\ &=6.25 \end{aligned}\)
and
\(\bar{X}_{2}=\frac{\sum_{i=1}^{8} X_{i}}{n_{2}}\)
\(=\frac{9+2+7+\ldots+8}{8}\)
\(=5.88\)
The sample variances are:
\(\begin{aligned} s_{1}^{2} &=\frac{\sum_{i=1}^{8}\left(X_{i}-\bar{X}_{1}\right)}{n_{1}-1} \\ &=\frac{(4-6.25)^{2}+(0-6.25)^{2}+\ldots+(5-6.25)^{2}}{8-1} \\ &=18.21 \end{aligned}\)
and
\(s_{2}^{2}=\frac{\sum_{i=1}^{8}\left(X_{i}-\bar{X}_{2}\right)}{n_{2}-1}\)
\(=\frac{(4-6.25)^{2}+(0-6.25)^{2}+\ldots+(5-6.25)^{2}}{8-1}\)
\(=9.84\)
The pooled variance estimator is:
\(s_{p}^{2}=\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}\)
\(=\frac{(8-1)(18.21)+(8-1)(9.84)}{8+8-2}\)
\(=14.03\)
The number of degrees of test statistic is:
\(\begin{aligned} v &=n_{1}+n_{2}-2 \\ &=8+8-2 \\ &=14 \end{aligned}\)
Now, substitute the values in the test statistic as shown below :
\(t=\frac{\left(\bar{x}_{1}-\bar{x}_{2}\right)-\left(\mu_{1}-\mu_{2}\right)}{\sqrt{s_{p}^{2}\left(\frac{1}{n_{1}}+\frac{1}{n_{2}}\right)}}\)
\(=\frac{(6.25-5.88)-(0)}{\sqrt{(14.03)\left(\frac{1}{8}+\frac{1}{8}\right)}}\)
\(=0.20025\)
The rejection region is:
\(t_{a, v}=t_{0.10,14}\)
\(=1.761\)
From the above, it concludes that the calculated value of \(t\) is less than the tabulated value of \(t\), so null hypothesis is accepted. Therefore, it infers that there is no difference in days absent between the two groups of workers.
After many years of teaching, a statistics professor computed the variance of the marks on her final exam and found it to be σ2 = 250. She recently made changes to the way in which the final exam is marked and wondered whether this would result in a reduction in the variance. A random sample of this year’s final exam marks are listed here. Can the professor infer at the 10% significance level that the variance has decreased?57 92 99 ...
A management professor was in the process of investigating the relationship between education and managerial level achieved. The source of his data was a survey of 385 CEOs of medium and large companies. He discovered that there was only one CEO who did not have at least one university degree. Estimate (using a Wilson estimator) with 99% confidence the proportion of CEOs of medium and large companies with no university degrees.
4. Comparing two population proportions (independent samples) Most major survey research organizations do not include wireless telephone numbers when conducting random-digit-dial telephone surveys. If there are differences between persons with and without landline phones, using a random-digit-dial telephone survey may introduce bias into the survey results. Data on a broad range of health topics are collected through personal household interviews of a representative sample of the U.S. civilian noninstitutionalized population in the National Health Interview Survey (NHIS) respondents are also asked about...
webwork / math243fall-masden /week 8b - ch20 inference about a population mean / 3 Week 8b - Ch20 Inference About a Population Mean: Problem 3 Previous Problem ListNext (1 point) An agricultural field trial compares the yield of two varieties of com. The researchers divide in half each of 10 fields of land in different locations and plarnt each corn variety in one half of each plot. After harvest, the yields are compared in bushels per acre at each location....
An independent-measures study comparing two treatment conditions produces a t statistic with df = 18. If the two samples are the same size, how many participants were in each of the samples? a) 9 b) 10 c) 19 d) 20
An Independent Samples T Test is used to make an inference about whether two groups in the population have different averages on some quantitative measure or dependent variable. This test could be used to compare the average annual earnings of High School graduates, compared to Community College graduates For this potential research question, answer the following questions. Please use only numeric values in your answers to questions 1, 2 and 3 and a word to answer question 4 1) According...
An Independent Samples T Test is used to make an inference about whether two groups in the population have different averages on some quantitative measure or dependent variable. This test could be used to compare the average annual earnings of High School graduates, compared to Community College graduates For this potential research question, answer the following questions. Please use only numeric values in your answers to questions 1, 2 and 3 and a word to answer question 4 1) According...
You are testing a claim about a population proportion of 0.2. Find the test statistic given the following sample results: successes: 70 sample size:225 Round to the fourth.
1. The T distribution can be use when comparing two population with population variance unknown. True False 2. You may use the Normal distribution when comparing two population with population variance known. True False