# Identify the null and alternative hypothesis in the following scenario. To determine if battery 1 lasts longer than battery 2, the mean lasting times, of the two competing batteries are compared. Twenty batteries of each type are randomly sampled and tested. Both populations have normal distributions with unknown standard deviations. Select the correct answer below: H_{0}:mu_{1}geqmu_{2}, H_{a}:mu_{1}−mu_{2} H_{0}:mu_{1}geq −mu_{2}, H_{a}:mu_{1}<−mu_{2} H_{0}:mu_{1}=mu_{2}, H_{a}:mu_{1}neq mu_{2} H_{0}:mu_{1}leq mu_{2}, H_{a}:mu_{1}>mu_{2}

Question
Normal distributions
Identify the null and alternative hypothesis in the following scenario.
To determine if battery 1 lasts longer than battery 2, the mean lasting times, of the two competing batteries are compared. Twenty batteries of each type are randomly sampled and tested. Both populations have normal distributions with unknown standard deviations.
Select the correct answer below: $$H_{0}:\mu_{1}\geq\mu_{2}, H_{a}:\mu_{1}<\mu_{2}$$</span>
$$H_{0}:\mu_{1}\leq −\mu_{2}, H_{a}:\mu_{1}>−\mu_{2}$$
$$H_{0}:\mu_{1}\geq −\mu_{2}, H_{a}:\mu_{1}<−\mu_{2}$$</span>
$$H_{0}:\mu_{1}=\mu_{2}, H_{a}:\mu_{1}\neq \mu_{2}$$
$$H_{0}:\mu_{1}\leq \mu_{2}, H_{a}:\mu_{1}>\mu_{2}$$

2021-01-20
Step 1: Given information
Twenty batteries of each type are randomly sampled and tested. Both populations have normal distributions with unknown standard deviations.
Step 2: Answer is given by,
Null hypothesis: $$H_{0} : \mu_{1} \leq \mu_{2}$$
Alternative hypothesis: $$H_{a} : \mu_{1} > \mu_{2}$$
Option (5) is correct.
Result: "$$H_{0}:\mu_{1}\leq \mu_{2},H_{a}:\mu_{1}>\mu_{2}$$"

### Relevant Questions

A random sample of $$\displaystyle{n}_{{1}}={16}$$ communities in western Kansas gave the following information for people under 25 years of age.
$$\displaystyle{X}_{{1}}:$$ Rate of hay fever per 1000 population for people under 25
$$\begin{array}{|c|c|} \hline 97 & 91 & 121 & 129 & 94 & 123 & 112 &93\\ \hline 125 & 95 & 125 & 117 & 97 & 122 & 127 & 88 \\ \hline \end{array}$$
A random sample of $$\displaystyle{n}_{{2}}={14}$$ regions in western Kansas gave the following information for people over 50 years old.
$$\displaystyle{X}_{{2}}:$$ Rate of hay fever per 1000 population for people over 50
$$\begin{array}{|c|c|} \hline 94 & 109 & 99 & 95 & 113 & 88 & 110\\ \hline 79 & 115 & 100 & 89 & 114 & 85 & 96\\ \hline \end{array}$$
(i) Use a calculator to calculate $$\displaystyle\overline{{x}}_{{1}},{s}_{{1}},\overline{{x}}_{{2}},{\quad\text{and}\quad}{s}_{{2}}.$$ (Round your answers to two decimal places.)
(ii) Assume that the hay fever rate in each age group has an approximately normal distribution. Do the data indicate that the age group over 50 has a lower rate of hay fever? Use $$\displaystyle\alpha={0.05}.$$
(a) What is the level of significance?
State the null and alternate hypotheses.
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}<\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}>\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}=\mu_{{2}},{H}_{{1}}:\mu_{{1}}\ne\mu_{{2}}$$
$$\displaystyle{H}_{{0}}:\mu_{{1}}>\mu_{{2}},{H}_{{1}}:\mu_{{1}}=\mu_{{12}}$$
(b) What sampling distribution will you use? What assumptions are you making?
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations,
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations,
The Student's t. We assume that both population distributions are approximately normal with known standard deviations,
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimalplaces.)
What is the value of the sample test statistic? (Test the difference $$\displaystyle\mu_{{1}}-\mu_{{2}}$$. Round your answer to three decimal places.)
(c) Find (or estimate) the P-value.
P-value $$\displaystyle>{0.250}$$
$$\displaystyle{0.125}<{P}-\text{value}<{0},{250}$$
$$\displaystyle{0},{050}<{P}-\text{value}<{0},{125}$$
$$\displaystyle{0},{025}<{P}-\text{value}<{0},{050}$$
$$\displaystyle{0},{005}<{P}-\text{value}<{0},{025}$$
P-value $$\displaystyle<{0.005}$$
Sketch the sampling distribution and show the area corresponding to the P-value.
P.vaiue Pevgiue
P-value f P-value
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.
Would you rather spend more federal taxes on art? Of a random sample of $$n_{1} = 86$$ politically conservative voters, $$r_{1} = 18$$ responded yes. Another random sample of $$n_{2} = 85$$ politically moderate voters showed that $$r_{2} = 21$$ responded yes. Does this information indicate that the population proportion of conservative voters inclined to spend more federal tax money on funding the arts is less than the proportion of moderate voters so inclined? Use $$\alpha = 0.05.$$ (a) State the null and alternate hypotheses. $$H_0:p_{1} = p_{2}, H_{1}:p_{1} > p_2$$
$$H_0:p_{1} = p_{2}, H_{1}:p_{1} < p_2$$
$$H_0:p_{1} = p_{2}, H_{1}:p_{1} \neq p_2$$
$$H_{0}:p_{1} < p_{2}, H_{1}:p_{1} = p_{2}$$ (b) What sampling distribution will you use? What assumptions are you making? The Student's t. The number of trials is sufficiently large. The standard normal. The number of trials is sufficiently large.The standard normal. We assume the population distributions are approximately normal. The Student's t. We assume the population distributions are approximately normal. (c)What is the value of the sample test statistic? (Test the difference $$p_{1} - p_{2}$$. Do not use rounded values. Round your final answer to two decimal places.) (d) Find (or estimate) the P-value. (Round your answer to four decimal places.) (e) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level alpha? At the $$\alpha = 0.05$$ level, we reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the $$\alpha = 0.05$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant. At the $$\alpha = 0.05$$ level, we reject the null hypothesis and conclude the data are not statistically significant. (f) Interpret your conclusion in the context of the application. Reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is sufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Fail to reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters. Reject the null hypothesis, there is insufficient evidence that the proportion of conservative voters favoring more tax dollars for the arts is less than the proportion of moderate voters.
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}&{H}{o}{u}{s}{e}{w}{\quad\text{or}\quad}{k}{H}{o}{u}{r}{s}\backslash{h}{l}\in{e}{G}{e}{n}{d}{e}{r}&{S}{a}\mp\le\ {S}{i}{z}{e}&{M}{e}{a}{n}&{S}{\tan{{d}}}{a}{r}{d}\ {D}{e}{v}{i}{a}{t}{i}{o}{n}\backslash{h}{l}\in{e}{W}{o}{m}{e}{n}&{473473}&{33.133}{.1}&{14.214}{.2}\backslash{h}{l}\in{e}{M}{e}{n}&{488488}&{18.618}{.6}&{15.715}{.7}\backslash{e}{n}{d}{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}$$ 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.
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.
Hypothesis Testing Review
For each problem below, simply identify the null and alternative hypotheses. Use appropriate notation/symbols. You do not have to run any hypothesis tests, although it's good practice and I'll post answers for all of them.
1) A simple random sample of 44 men from a normally distributed population results in a standard deviation of 10.7 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.10 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute.
2) In 1997, a survey of 880 households showed that 145 of them use e-mail. Use those sample results to test the claim that more than 15% of households use e-mail. Use a 0.05 significance level.
Hypothesis Testing Review
For each problem below, simply identify the null and alternative hypotheses. Use appropriate notation/symbols. You do not have to run any hypothesis tests, although it's good practice and I'll post answers for all of them.
1) A simple random sample of 44 men from a normally distributed population results in a standard deviation of 10.7 beats per minute. The normal range of pulse rates of adults is typically given as 60 to 100 beats per minute. If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute. Use the sample results with a 0.10 significance level to test the claim that pulse rates of men have a standard deviation equal to 10 beats per minute.
2) In 1997, a survey of 880 households showed that 145 of them use e-mail. Use those sample results to test the claim that more than 15% of households use e-mail. Use a 0.05 significance level.
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.
For a two-tailed hypothesis test with level of significance a and null hypothesis $$H_{0} : \mu = k$$ we reject Ho whenever k falls outside the $$c = 1 — \alpha$$ confidence interval for $$\mu$$ based on the sample data. When A falls within the $$c = 1 — \alpha$$ confidence interval. we do reject $$H_{0}$$.
For a one-tailed hypothesis test with level of significance Ho : $$\mu = k$$ and null hypothesiswe reject Ho whenever A falls outsidethe $$c = 1 — 2\alpha$$ confidence interval for p based on the sample data. When A falls within the $$c = 1 — 2\alpha$$ confidence interval, we do not reject $$H_{0}$$.
A corresponding relationship between confidence intervals and two-tailed hypothesis tests is also valid for other parameters, such as p, $$\mu1 — \mu_2,\ and\ p_{1}, - p_{2}$$.
(a) Consider the hypotheses $$H_{0} : \mu_{1} — \mu_{2} = O\ and\ H_{1} : \mu_{1} — \mu_{2} \neq$$ Suppose a 95% confidence interval for $$\mu_{1} — \mu_{2}$$ contains only positive numbers. Should you reject the null hypothesis when $$\alpha = 0.05$$? Why or why not?