Fill in the blank to finish the statement.If we perform not reject the null hypothesis as soon as the explain in the alternative hypothesis is​ true, we have made a Type​ _______ error.

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A form II error occurs if the null hypothesis is not rejected​ when, in​ fact, the different hypothesis is true.
The null and alternate hypotheses room given. Recognize whether the hypothesis test is​ left-tailed, right-tailed, or​ two-tailed. What parameter is being​ tested?H0​: σ=110H1​: σ
(a) determine the null and also alternative​ hypotheses, (b) explain what the would typical to make a kind I​ error, and​ (c) explain what that would median to make a type II error.Three years​ ago, the average price that a​ single-family house was ​\$243,782. A actual estate broker believes that the mean price has actually increased due to the fact that then.(a) i beg your pardon of the following is the hypothesis test to be conducted?(b) i m sorry of the adhering to is a form I error?(c) i m sorry of the following is a type II error?
(a) H0: μ=\$243,782; H1: μ>\$243,782.(b) The broker rejects the theory that the median price is \$243,782, once it is the true median cost.(c) The broker fails to reject the theory that the typical price is \$243,782, once the true mean price is greater than \$243,782.
Five years​ ago, 10.5​% of high college students had actually tried marijuana for the very first time prior to the age of 13. A school resource officer​ (SRO) thinks the the relationship of high college students who have tried marijuana because that the an initial time prior to the age of 13 has increased since then.​(a) identify the null and alternate hypotheses. Which of the adhering to is​ correct?(b) intend sample data suggest that the null theory shouldbe rejected. State the conclusion of the researcher.(c) Suppose, in​ fact, the the proportion of high school students who have tried marijuana prior to the period of 13 to be 10.5​%. Was a form I or type II error​ committed?
(a) H0: p=0.105; H1:p>0.105(b) there is sufficient evidence come conclude that the proportion of high college students has actually increased.(c) The SRO cursed a kind IIerror since he garbage the null hypothesis when, infact, it is true.
Determine if the following statement is true or false.When trial and error a hypothesis utilizing the​ P-value Approach, if the​ P-value is​ large, reject the null hypothesis.
Test the hypothesis utilizing the​ P-value approach. Be certain to verify the demands of the test.H0​: p=0.8 versus H1​: p>0.8n=125; x=115; α=0.05(a) Is np0 (1-p0)≥​10?(b) What is the P-value?
In a clinical​ trial, 31 out of 900patients acquisition a prescription medicine complained of flulike symptoms. Intend that the is known that 2.3​% of patient taking completing drugs complain of flulike symptoms. Is there adequate evidence to conclude that an ext than 2.3​% of this​ drug"s users experience flulike symptoms together a side result at the α=0.05 level of​ significance?(a) What are the null and also alternative​ hypotheses?(b) What is the P-value?
(a) H0: p=0.023 versus H1: p>0.023(b) 0.014; due to the fact that ​P-valuethe null hypothesis and conclude the there is enough evidence that much more than 2.3​% the the users experience flulike symptoms.
In a recent​ survey, 37​% of work U.S. Adult reported that an easy mathematical skills were critical or very important to their job. The supervisor of the job placement office at a​ 4-year college thinks this percentage has actually increased due to increased use of technology in the workplace. She bring away a random sample the 800employed adults and finds that 321of them feeling that an easy mathematical skills are crucial or really important to your job. Is there enough evidence to conclude that the percent of to work adults who feel an easy mathematical an abilities are an important or an extremely important to your job has increased in ~ the α=0.01 level of​ significance?(a) What space the null and alternate hypotheses?(b) identify the test statistic?(c)
In a previous​ poll, 49​% of adults with youngsters under the period of 18 reported that their family members ate dinner with each other 7 nights a week. Suppose​ that, in a more recent​ poll, 554 the 1150 adults with children under the age of 18 reported that their family members ate dinner with each other 7 nights a week. Is there enough evidence the the relationship of family members with kids under the age of 18 that eat dinner together 7 nights a week has reduced at the α=0.05 significance​ level?(a) What room the null and alternative​ hypotheses?(b) What is the P-value?(c)
(a) H0: p=0.36 versus H1: p(b) P-value=0.064; ​Yes, over there is enough evidence due to the fact that the​ P-value is lessthan the level that significance.​Therefore, rejectthe null hypothesis.
​Previously, 35​% of parental of youngsters in high institution felt it was a serious difficulty that high school students were no being taught sufficient math and also science. A current survey uncovered that 120 the 300 parents of kids in high college felt it to be a serious trouble that high college students were not being taught enough math and science. Do parents feel differently​ today? usage the α=0.05level that significance.(a) What are the null and also alternative​ hypotheses?(b) determine the test statistic?(c) What is the P-value?(d) analyze the P-value.(e) identify the conclusion because that this hypothesis test.
(a) H0: p=0.48 versus H1: p≠0.48(b) Z0=1.27(c) P-value=0.204(d) The​ P-value is the probability the observing a sample statistic as excessive or more extreme 보다 the one derived if the population proportion amounts to 0.48.(e) since P-value>α​,do not reject the null hypothesis and conclude the there is not adequate evidence the parents feel in different way today.
In a previous​ year, 65​% that females aged 15 year of age and older live alone. A sociologist tests even if it is this percentage is different today by conducting a arbitrarily sample the 700 females age 15 year of age and older and finds that 462 are living alone. Is there sufficient evidence in ~ the α=0.05 level of significance to break up the proportion has​ changed?(a) recognize the null and alternative hypotheses because that this test.(b) find the check statistic because that this theory test.(c) recognize the P-value because that this hypothesis test.(d) State the conclusion because that this theory test.
(a) H0: p=0.63; H1: p≠0.63(b) 1.52(c) 0.129(d) perform not disapprove H0. There is not enough evidence at the α​=0.05 level of meaning to conclude that the proportion of females who space living alone has actually changed.
To check H0: μ=20 matches H1: μ(a) If x overbar = 18.3 and s = 4.4, compute the test statistic.(b) draw a​ t-distribution v the area that represents the​ P-value shaded. I m sorry of the following graphs reflects the exactly shaded​ region?(c) almost right the​ P-value. Choose the correct selection for the​ P-value below.(d) If the researcher decides to check this theory at the α=0.05 level of​ significance, will the researcher refuse the null​ hypothesis?
(a) t=-1.55(b) Graph shown.(c) 0.05 (d) The researcher will not refuse the null hypothesis due to the fact that the P-value is not much less than α.
To test H0: μ=80 versus H1: μ≠80, a basic random sample of size n = 26 is acquired from a population that is well-known to be generally distributed. Complete parts (a) with (c) below.(a) If x overbar = 83.1 and s = 14.3, compute the test statistic. (b) suppose a researcher wants to test this hypothesis at a level of significance of α=0.02. Approximate the​ P-value using the​ t-distribution table.
(a) t=1.105(b) 0.20 ≤ P-value ≤ 0.30(c) No​, the researcher will not disapprove the null hypothesis since the​ p-value is greatergreaterthan the level that significance.
Several years​ ago, the mean elevation of women two decades of age or older was 63.7 inches. Intend that a random sample of 45 women that are twenty years of period or larger today results in a mean elevation of 64.9 inches.​(a) State the proper null and alternative hypotheses to assess whether women space taller today.​(b) mean the​ P-value because that this check is 0.16. Explain what this worth represents.​(c) create a conclusion for this hypothesis test assuming an α=0.10 level the significance.
(a) H0: μ=63.7 in. Versus H1: μ>63.7 in.(b) there is a 0.16probability the obtaining a sample mean height of 64.9inches or higher from a populace whose mean height is 63.7inches.(c) do not disapprove the null hypothesis. There is not enough evidence come conclude the the mean height of women twenty years of age or larger is better today.
The median waiting time in ~ the​ drive-through that a​ fast-food restaurant from the time an stimulate is inserted to the moment the stimulate is received is 86.1 seconds. A manager devises a new​ drive-through system that she believes will certainly decrease wait time. As a​ test, she initiates the new system at herrestaurant and also measures the wait time because that 10 randomly selected orders. The wait times are noted in the table to the right. Complete parts​ (a) and​ (b) below.(a) since the sample size is​ small, the manager need to verify the the wait time is typically distributed and the sample does no contain any kind of outliers. The regular probability plot is presented below and also the sample correlation coefficient is recognized to be r=0.987.Are the conditions for experimentation the hypothesis​ satisfied?(b) Is the new system​ effective? command a theory test utilizing the​ P-value approach and a level of meaning of α=0.05.First recognize the appropriate hypothesis.Find the check statistic.Find the P-value.Use the α=0.05 level of significance. What deserve to be concluded from the hypothesis test?
(a) Yes, the conditions are satisfied. The regular probability plot is direct enough, due to the fact that the correlation coefficient is better than the an essential value. In addition, a box plot does not show any kind of outliers.(b) H0: μ=86.1H1: μt=-1.58P-value=0.074The P-value is better than the level of meaning so over there is not adequate evidence come conclude the new system is effective.
A golf association needs that golf balls have a diameter that is 1.68 inches. To recognize if golf balls conform to the​ standard, a arbitrarily sample the golf balls to be selected. Their diameters are displayed in the data table. Finish parts​ (a) and​ (b) below.(a) since the sample dimension is​ small, the engineer should verify the the diameter is typically distributed and also the sample does no contain any type of outliers. The normal probability plot is shown below and also the sample correlation coefficient is known to it is in r=0.952. Are the conditions for experimentation the hypothesis​ satisfied?(b) carry out the golf balls conform to the​ standards? conduct a theory test using the​ P-value approach and a level of meaning of α=0.05.First recognize the appropriate hypothesis.Find the check statistic.Find the P-value.Use the α=0.05 level of significance. What have the right to be concluded from the hypothesis test?
(a) Yes, the conditions are satisfied. The regular probability plot is direct enough, due to the fact that the correlation coefficient is greater than the critical value.(b) H0: μ=1.68H1: μ≠1.68t=0.75P-value=0.468There is not sufficient evidence to conclude the the golf balls do not conform come the association"s standards.
Real estate taxes are supplied to fund local education​ institutions, police​ protection, libraries, and also other neighborhood public services. A​ household"s genuine estate taxation bill is based upon the worth of the house. In a recent​ year, the typical real estate taxes bill per​ \$1,000 assessed value for all families in a certain country was \$8.43. A random sample that 50 rural families results in a sample typical real estate bill per​ \$1,000 assessed value of \$6.57 with a traditional deviation the \$5.98.Use the level of definition α=0.05. Complete parts​ (a) and​ (b).​(a) based upon the histogram and boxplot​ shown, why is a big sample important to command a hypothesis test around the​ mean?(b) conduct a hypothesis test utilizing the​ P-value approach and a level of significance of α=0.05to identify if the evidence says the average real estate invoice for rural households per​ \$1,000 assessed value differs from the of all households.First identify the appropriate hypothesis.Find the check statistic.Find the P-value.Use the α=0.05 level the significance. What deserve to be concluded from the theory test?
(a) The real estate tax bill circulation is skewed best with outliers.(b) H0: μ=8.43H1: μ≠8.43t=-2.20P-value=0.033The evidence suggests that rural households have a various mean actual estate taxes bill.
Calcium is essential to tree growth because it disclosure the formation of wood and also maintains cabinet walls. In​ 1990, the concentration of calcium in precipitation in a specific area to be 0.12 milligrams per liter​ (mg/L). A arbitrarily sample the 10 precipitation days in 2007 results in the adhering to data table. Complete parts​ (a) through​ (c) below.(a) State the hypotheses for determining if the typical concentration the calcium precipitation has readjusted since 1990.(b) build a 95​% to trust interval around the sample median concentration the calcium precipitation.(c) does the sample evidence suggest the calcium concentration have changed since​ 1990?
(a) H0: μ=0.12 mg/LH1: μ≠0.12 mg/L(b) The lower bound is 0.0931.The top bound is 0.2183.(c) No, because the confidence interval has 0.12 mg/L.
The average daily volume the a computer stock in 2011 to be μ=35.1million​ shares, according to a reputable source. A stock analyst believes the the share volume in 2014 is different from the 2011 level. Based upon a random sample that 30 trading work in​ 2014, he find the sample typical to it is in 29.8million​ shares, with a traditional deviation the s=14.4 million shares. Test the hypotheses by creating a 95​%confidence interval. Finish parts​ (a) through​ (c) below.(a) State the hypotheses because that the test.(b) construct a 95​% confidence interval around the sample median of share traded in 2014.(c) will the researcher reject the null​ hypothesis?
(a) H0: μ=35.1 million shares.H1: μ≠35.1 million shares(b) The reduced bound is 24.423 million shares.The upper bound is 35.177 million shares.(c) perform not disapprove the null hypothesis because μ=35.1 million shares drops in the confidence interval.

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It has actually long been stated that the typical temperature of people is 98.6°F. However, two researchers currently involved in the topic thought the the median temperature of people is less than 98.6°F. They measured the temperatures of 56 healthy and balanced adults 1 come 4 times day-to-day for 3​ days, obtaining 250 measurements. The sample data led to a sample median of 98.2°F and also a sample standard deviation of 11°F. Use the​ P-value technique to command a hypothesis test to judge whether the typical temperature of humans is less than 98.6°F at the α=0.01level the significance.(a) State the hypotheses.(b) find the test statistic.(c) find the P-value.(d) What deserve to be concluded?
(a) H0: μ=98.6°FH1: μ(b) t=-6.32(c) P-value=0.000(d) reject H0 because the P-value is less than the significance level.
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