# Consider two independent populations that are normally distributions. A simple random sample of n_1=41 from the first population showed bar x_11=33, and a simple random of size n_2=48 from the second population showed bar x_2=32 Suppose s_1=9s_1=9 and s_2=10s_2=10, find a 98% confidence interval for mu_1−mu_2mu_1-mu_2. (Round answers to two decimal places.) margin of error-? lower limit-? upper limit-?

Question
Normal distributions
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-?

2021-03-12
Step 1
From the provided information,
$$\displaystyle{n}_{{1}}={41}$$
$$\displaystyle\overline{{x}}_{{1}}={33}$$
$$\displaystyle{n}_{{2}}={48}$$
$$\displaystyle\overline{{x}}_{{2}}={32}$$
$$\displaystyle{s}_{{1}}={9}$$
$$\displaystyle{s}_{{2}}={10}$$
The confidence level = 98%
Level of significance $$\displaystyle{\left(\alpha\right)}={1}-{0.98}={0.02}$$
The degree of freedom $$\displaystyle={n}_{{1}}+{n}_{{2}}-{2}={41}+{48}-{2}={87}$$
The critical value of t at 87 degree of freedom with 0.02 level of significance from the t value table is 2.37.
Step 2
The pooled standard deviation can be obtained as:
$$\displaystyle{s}_{{p}}=\sqrt{{\frac{{{\left({n}_{{1}}-{1}\right)}{{s}_{{1}}^{{2}}}+{\left({n}_{{2}}-{1}\right)}{{s}_{{2}}^{{2}}}}}{{{n}_{{1}}+{n}_{{2}}-{2}}}}}$$
$$\displaystyle=\sqrt{{\frac{{{\left({41}-{1}\right)}{\left({9}\right)}^{{2}}+{\left({48}-{1}\right)}{\left({10}\right)}^{{2}}}}{{{41}+{48}-{2}}}}}$$
=9.553
The required margin of error can be obtained as:
$$\displaystyle{E}={t}{\left({s}_{{p}}\sqrt{{\frac{{1}}{{{n}_{{1}}}}+\frac{{1}}{{{n}_{{2}}}}}}\right)}$$
$$\displaystyle={\left({2.37}\right)}{\left({9.553}\sqrt{{\frac{{1}}{{41}}+\frac{{1}}{{48}}}}\right)}$$
$$\displaystyle\approx{4.81}$$
Thus, the margin of error is 4.81.
Step 3
The required 98% confidence interval for $$\displaystyle\mu_{{1}}-\mu_{{2}}$$ can be obtained as:
$$\displaystyle{C}{I}={\left(\overline{{x}}_{{1}}-\overline{{x}}_{{1}}\right)}\pm{E}$$
$$\displaystyle={\left({33}-{32}\right)}\pm{4.81}$$
=1+-4.81
=(-3.81,5.81)
Thus, the lower limit of confidence interval is -3.81 and the upper limit is 5.81.

### Relevant Questions

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.
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 $$\displaystyle\mu_{{1}}−\mu_{{2}}$$.
A random sample of 5805 physicians in Colorado showed that 3332 provided at least some charity care (i.e., treated poor people at no cost).
a)
Let p represent the proportion of all Colorado physicians who provide some charity care. Find a point estimate for p. (Round your answer to four decimal places.)
b)
Find a $$99\%$$ confidence interval for p. (Round your answer to three decimal places.)
lower limit $$\displaystyle=?$$
upper limit $$\displaystyle=?$$
c)
Is the normal approximation to the binomial justified in this problem? Explain
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.
A machine produces parts with lengths that are normally distributed with $$\displaystyle\sigma={0.66}$$. A sample of 20 parts has a mean length of 75.98.
a)
Give a point estimate for $$\displaystyle\mu$$. (Give your answer correct to two decimal places.)
b)
Find the $$95\%$$ confidence maximum error of estimate for $$\displaystyle\mu$$. (Give your answer correct to two decimal places.)
c)
Find the $$95\%$$ confidence interval for $$\displaystyle\mu$$. (Give your answer correct to two decimal places.)
Lower limit $$\displaystyle=?$$
Upper limit $$\displaystyle=?$$

Assume that the random variable Z follows standard normal distribution, calculate the following probabilities (Round to two decimal places)
a)$$P(z>1.9)$$
b)$$\displaystyle{P}{\left(−{2}\le{z}\le{1.2}\right)}$$
c)$$P(z\geq0.2)$$

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.)
In 2014, the Centers for Disearse reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of 0.30.
a) How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of 0.02? Use 95% confidence.(Round your answer up to the nearest integer.)
b)Assume that the study uses your sample size recommendation in part (a) and finds 470 smokers. What is the point estimate of the proportion of smokers in the population? (Round your answer to four decimal places.)
c) What is the 95% confidence interval for the proportion of smokers in the population? (Round your answer to four decimal places.)
When we want to test a claim about two population means, most of the time we do not know the population standard deviations, and we assume they are not equal. When this is the case, which of the following is/are not true?
-The samples are dependent
-The two populations have to have uniform distributions
-Both samples are simple random samples
-Either the two sample sizes are large or both samples come from populations having normal distributions or both of these conditions satisfied.
You intend to estimate a population mean $$\displaystyle\mu$$ with the following sample.
You believe the population is normally distributed. Find the $$99.9\%$$ confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to twp decimal places (because the sample data are reported accurate to one decimal place).
$$\displaystyle{99.9}\%{C}.{I}.={\left({35.68},{61.54}\right)}$$Incorrect