# Let two independent random samples, each of size 10, from two normal distributions N(mu_1, sigma_2) and N(mu_2, sigma_2) yield x = 4.8, s_1^2 = 8.64, y = 5.6, s_2^2 = 7.88. Find a 95% confidence interval for mu_1 − mu_2.

Question
Normal distributions
Let two independent random samples, each of size 10, from two normal distributions $$\displaystyle{N}{\left(\mu_{{1}},\sigma_{{2}}\right)}{\quad\text{and}\quad}{N}{\left(\mu_{{2}},\sigma_{{2}}\right)}$$ yield $$\displaystyle{x}={4.8},{{s}_{{1}}^{{2}}}$$
$$\displaystyle={8.64},{y}={5.6},{{s}_{{2}}^{{2}}}$$
= 7.88.
Find a 95% confidence interval for $$\displaystyle\mu_{{1}}−\mu_{{2}}$$.

2021-02-05
Step 1
$$\displaystyle{A}{100}{\left({1}−\alpha\right)}\%$$ confidence interval on the difference is
$$\displaystyle\overline{{x}}_{{1}}-\overline{{x}}_{{2}}-{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\le\mu_{{1}}-\mu_{{2}}\le\overline{{x}}_{{1}}-\overline{{x}}_{{2}},+{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}$$
where $$\displaystyle{s}_{{p}}=\sqrt{{\frac{{{\left({n}_{{1}}-{1}\right)}{{s}_{{1}}^{{2}}}+{\left({n}_{{2}}-{1}\right)}{{s}_{{2}}^{{2}}}}}{{{n}_{{1}}+{n}_{{2}}-{1}}}}}$$
Step 2
Considering 95% confidence interval,
$$\displaystyle\alpha={0.05}$$
$$\displaystyle\overline{{x}}_{{1}}={4.8}$$
$$\displaystyle\overline{{x}}_{{2}}={5.6}$$
$$\displaystyle{{s}_{{1}}^{{2}}}={8.64}$$
$$\displaystyle{{s}_{{2}}^{{2}}}={7.88}$$
$$\displaystyle{n}_{{1}}={10}$$
$$\displaystyle{n}_{{2}}={10}$$
Therefore
$$\displaystyle{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}={t}_{{\frac{{0.05}}{{2}},{10}+{10}-{1}}}$$
$$\displaystyle={t}_{{{0.025},{19}}}$$
=2.093
Step 3
To calculate $$\displaystyle{s}_{{p}}$$
$$\displaystyle{s}_{{p}}=\sqrt{{\frac{{{\left({n}_{{1}}-{1}\right)}{{s}_{{1}}^{{2}}}+{\left({n}_{{2}}-{1}\right)}{{s}_{{2}}^{{2}}}}}{{{n}_{{1}}+{n}_{{2}}-{1}}}}}$$
$$\displaystyle{s}_{{p}}=\frac{\sqrt{{{\left({10}-{1}\right)}{8.64}+{\left({10}-{1}\right)}{7.88}}}}{{{19}}}{)}$$
$$\displaystyle{s}_{{p}}={2.797}$$
Step 4
Now, by substituting all the values in the equation, we get
$$\displaystyle\overline{{x}}_{{1}}-\overline{{x}}_{{2}}-{t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\le\mu_{{1}}-\mu_{{2}}\le\overline{{x}}_{{1}}-\overline{{x}}_{{2}}$$
$$\displaystyle={t}_{{\frac{\alpha}{{2}},{n}_{{1}}+{n}_{{2}}-{1}}}{s}{p}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}$$
$$\displaystyle{4.8}-{5.6}-{2.093}\times{2.797}\times{0.447}\le\mu_{{1}}-\mu_{{2}}\le{4.8}-{5.6}+{2.093}\times{2.797}\times{0.447}-{3.424}\le\mu_{{1}}-\mu_{{2}}\le{1.82}$$
Hence 95% confidence interval in $$\displaystyle\mu_{{1}}-\mu_{{2}}{i}{s}-{3.424}\le\mu_{{1}}-\mu_{{2}}\le{1.82}$$

### Relevant Questions

Consider two independent populations that are normally distributions. A simple random sample of $$\displaystyle{n}_{{1}}={41}$$ from the first population showed $$\displaystyle\overline{{x}}_{{11}}={33}$$, and a simple random of size $$\displaystyle{n}_{{2}}={48}$$ from the second population showed
$$\displaystyle\overline{{x}}_{{2}}={32}$$
Suppose $$\displaystyle{s}_{{1}}={9}{s}_{{1}}={9}{\quad\text{and}\quad}{s}_{{2}}={10}{s}_{{2}}={10}$$, find a 98% confidence interval for $$\displaystyle\mu_{{1}}−\mu_{{2}}\mu_{{1}}-\mu_{{2}}$$. (Round answers to two decimal places.)
margin of error-?
lower limit-?
upper limit-?
Suppose you take independent random samples from populations with means $$\displaystyle\mu{1}{\quad\text{and}\quad}\mu{2}$$ and standard deviations $$\displaystyle\sigma{1}{\quad\text{and}\quad}\sigma{2}$$. Furthermore, assume either that (i) both populations have normal distributions, or (ii) the sample sizes (n1 and n2) are large. If X1 and X2 are the random sample means, then how does the quantity
$$\displaystyle\frac{{{\left(\overline{{{x}_{{1}}}}-\overline{{{x}_{{2}}}}\right)}-{\left(\mu_{{1}}-\mu_{{2}}\right)}}}{{\sqrt{{\frac{{{\sigma_{{1}}^{{2}}}}}{{{n}_{{1}}}}+\frac{{{\sigma_{{2}}^{{2}}}}}{{{n}_{{2}}}}}}}}$$
Give the name of the distribution and any parameters needed to describe it.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}&{H}{o}{u}{s}{e}{w}{\quad\text{or}\quad}{k}{H}{o}{u}{r}{s}\backslash{h}{l}\in{e}{G}{e}{n}{d}{e}{r}&{S}{a}\mp\le\ {S}{i}{z}{e}&{M}{e}{a}{n}&{S}{\tan{{d}}}{a}{r}{d}\ {D}{e}{v}{i}{a}{t}{i}{o}{n}\backslash{h}{l}\in{e}{W}{o}{m}{e}{n}&{473473}&{33.133}{.1}&{14.214}{.2}\backslash{h}{l}\in{e}{M}{e}{n}&{488488}&{18.618}{.6}&{15.715}{.7}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ a. Based on this​ study, calculate how many more hours per​ week, on the​ average, women spend on housework than men. b. Find the standard error for comparing the means. What factor causes the standard error to be small compared to the sample standard deviations for the two​ groups? The cause the standard error to be small compared to the sample standard deviations for the two groups. c. Calculate the​ 95% confidence interval comparing the population means for women Interpret the result including the relevance of 0 being within the interval or not. The​ 95% confidence interval for ​$$\displaystyle{\left(\mu_{{W}}-\mu_{{M}}​\right)}$$ is: (Round to two decimal places as​ needed.) The values in the​ 95% confidence interval are less than 0, are greater than 0, include 0, which implies that the population mean for women could be the same as is less than is greater than the population mean for men. d. State the assumptions upon which the interval in part c is based. Upon which assumptions below is the interval​ based? Select all that apply. A.The standard deviations of the two populations are approximately equal. B.The population distribution for each group is approximately normal. C.The samples from the two groups are independent. D.The samples from the two groups are random.
A two-sample inference deals with dependent and independent inferences. In a two-sample hypothesis testing problem, underlying parameters of two different populations are compared. In a longitudinal (or follow-up) study, the same group of people is followed over time. Two samples are said to be paired when each data point in the first sample is matched and related to a unique data point in the second sample.
This problem demonstrates inference from two dependent (follow-up) samples using the data from the hypothetical study of new cases of tuberculosis (TB) before and after the vaccination was done in several geographical areas in a country in sub-Saharan Africa. Conclusion about the null hypothesis is to note the difference between samples.
The problem that demonstrates inference from two dependent samples uses hypothetical data from the TB vaccinations and the number of new cases before and after vaccination. PSK\begin{array}{|c|c|} \hline Geographical\ regions & Before\ vaccination & After\ vaccination\\ \hline 1 & 85 & 11\\ \hline 2 & 77 & 5\\ \hline 3 & 110 & 14\\ \hline 4 & 65 & 12\\ \hline 5 & 81 & 10\\\hline 6 & 70 & 7\\ \hline 7 & 74 & 8\\ \hline 8 & 84 & 11\\ \hline 9 & 90 & 9\\ \hline 10 & 95 & 8\\ \hline \end{array}ZSK
Using the Minitab statistical analysis program to enter the data and perform the analysis, complete the following: Construct a one-sided $$\displaystyle{95}\%$$ confidence interval for the true difference in population means. Test the null hypothesis that the population means are identical at the 0.05 level of significance.
Are yields for organic farming different from conventional farming yields? Independent random samples from method A (organic farming) and method B (conventional farming) gave the following information about yield of sweet corn (in tons/acre). $$\text{Method} A: 6.51, 7.02, 6.81, 7.27, 6.73, 6.11, 6.17, 5.88, 6.69, 7.12, 5.74, 6.90.$$
$$\text{Method} B: 7.32, 7.01, 6.66, 6.85, 5.78, 6.48, 5.95, 6.31, 6.50, 5.93, 6.68.$$ Use a 5% level of significance to test the claim that there is no difference between the yield distributions. (a) What is the level of significance? (b) Compute the sample test statistic. (Round your answer to two decimal places.) (c) Find the P-value of the sample test statistic. (Round your answer to four decimal places.)
A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of $$25^{\circ}F$$. However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to $$25^{\circ}F$$. One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of 5.1. Another similar frozen food case was equipped with the old thermostat, and a random sample of 19 temperature readings gave a sample variance of 12.8. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a $$5\%$$ level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)
(a) What is the level of significance?
State the null and alternate hypotheses.
$$H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}>?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}\neq?_{2}^{2}H0:?_{1}^{2}=?_{2}^{2},H1:?_{1}^{2}?_{2}^{2},H1:?_{1}^{2}=?_{2}^{2}$$
(b) Find the value of the sample F statistic. (Round your answer to two decimal places.)
What are the degrees of freedom?
$$df_{N} = ?$$
$$df_{D} = ?$$
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.
Assume two normal distributions where $$\displaystyle\mu_{{1}}={0.0001},\sigma_{{1}}={0.01},\mu_{{2}}=-{.0002},\sigma_{{2}}={0.015},{\quad\text{and}\quad}ρ={.45}$$. Using $$\displaystyle{z}_{{s}}={\left\lbrace{.814},{.259}\right\rbrace}$$, generate $$\displaystyle{z}_{{1}}{\quad\text{and}\quad}{z}_{{2}}$$.
a. $$\displaystyle\alpha={0.05},{n}_{{1}}={30},{s}_{{1}}={16.37},{n}_{{2}}={39},{s}_{{2}}={9.88},$$
b. $$\displaystyle\alpha={0.01},{n}_{{1}}={25},{s}_{{1}}={5.2},{n}_{{2}}={20},{s}_{{2}}={6.8}$$
At what age do babies learn to crawl? Does it take longer to learn in the winter when babies are often bundled in clothes that restrict their movement? Data were collected from parents who brought their babies into the University of Denver Infant Study Center to participate in one of a number of experiments between 1988 and 1991. Parents reported the birth month and the age at which their child was first able to creep or crawl a distance of 4 feet within 1 minute. The resulting data were grouped by month of birth: January, May, and September: $$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}&{C}{r}{a}{w}{l}\in{g}\ {a}\ge\backslash{h}{l}\in{e}{B}{i}{r}{t}{h}\ {m}{o}{n}{t}{h}&{M}{e}{a}{n}&{S}{t}.{d}{e}{v}.&{n}\backslash{h}{l}\in{e}{J}{a}\nu{a}{r}{y}&{29.84}&{7.08}&{32}\backslash{M}{a}{y}&{28.58}&{8.07}&{27}\backslash{S}{e}{p}{t}{e}{m}{b}{e}{r}&{33.83}&{6.93}&{38}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ Crawling age is given in weeks. Assume the data represent three independent simple random samples, one from each of the three populations consisting of babies born in that particular month, and that the populations of crawling ages have Normal distributions. A partial ANOVA table is given below. $$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{c}\right\rbrace}{S}{o}{u}{r}{c}{e}&{S}{u}{m}\ {o}{f}\ \boxempty{s}&{D}{F}&{M}{e}{a}{n}\ \boxempty\ {F}\backslash{h}{l}\in{e}{G}{r}{o}{u}{p}{s}&{505.26}\backslash{E}{r}{r}{\quad\text{or}\quad}&&&{53.45}\backslash{T}{o}{t}{a}{l}{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ What are the degrees of freedom for the groups term?
Calculate confidence intervals for ratio of two population variances and ratio of standard deviations. Assume that samples are simple random samples and taken from normal populations. a) $$\displaystyle\alpha={0.05},\ {n}_{{{1}}}={30},\ {s}_{{{1}}}={16.37},\ {n}_{{{2}}}={39},\ {s}_{{{2}}}={9.88}$$ b) $$\displaystyle\alpha={0.01},\ {n}_{{{1}}}={25},\ {s}_{{{1}}}={5.2},\ {n}_{{{2}}}={20},\ {s}_{{{2}}}={6.8}$$