STAT CH 7

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Find the critical​ value(s) and rejection​ region(s) for the indicated​ t-test, level of significance alpha​, and sample size n. Left​-tailed ​test, alpha = 0.05​, n = 21

(use t-dist table) The critical​ value(s) is/are -2.845. ^To find the critical values in a​ t-distribution, first identify the level of​ significance, alpha​, and the degrees of​ freedom, d.f. = n-1. Since this is a​ left-tailed test, use the​ one-tail column value of the​ t-distribution table, with a negative sign. If this is a​ two-tailed test, there will be two critical values. Use the​ two-tails column value with a negative sign for one critical value and a positive sign for the other critical value. (-2.845, 2.845) Determine the rejection​ region(s). t < -2.845 ^ If it is a​ left-tailed test, then the rejection region is to the left of the critical value. If it is a​ right-tailed test, then the rejection region is to the right of the critical value. If it is a​ two-tailed test, then the rejection regions are to the left of the negative critical value and to the right of the positive critical value.

What type of test is being conducted?

= : two- tailed test >= : left- tailed test <= : right- tailed test

What are the two decisions that you can make from performing a hypothesis​ test? Select all that apply.

A. accept the null hypothesis B. make a type I error C. fail to reject the null hypothesis >D. reject the alternative hypothesis >E. reject the null hypothesis F. accept the alternative hypothesis G. fail to reject the alternative hypothesis H. make a type II error

Determine whether the claim stated below represents the null hypothesis or the alternative hypothesis. If a hypothesis test is​ performed, how should you interpret a decision that​ (a) rejects the null hypothesis or​ (b) fails to reject the null​ hypothesis? A scientist claims that the mean incubation period for the eggs of a species of bird is more than 40 days.

Does the claim represent the null hypothesis or the alternative​ hypothesis? Since the claim does not contain a statement of​ equality, it represents the alternative hypothesis. ​(a) How should you interpret a decision that rejects the null​ hypothesis? There is sufficient evidence to support the claim that the mean incubation period for the eggs of a species of bird is more than 40 days. ​ (b) How should you interpret a decision that fails to reject the null​ hypothesis? There is insufficient evidence to support the claim that the mean incubation period for the eggs of a species of bird is more than 40 days.

Use the given statement to represent a claim. Write its complement and state which is (Upper) H0 and which is Ha. mu greater than or equals 637.

Find the complement of the claim. mu < 637 Which is Upper H 0 and which is Ha​? Upper H 0​: mu > or = 637 Ha​: mu < 637

Find the critical​ value(s) and rejection​ region(s) for a​ two-tailed chi-square test with a sample size n = 13 and level of significance alpha = 0.10.

Find the critical​ value(s). 5.226, 21.026 ^ 2 tails = 1/2 alpha, d.f. = n-1

Explain how to find the critical values for a​ t-distribution.

Give the first step. Choose the correct answer below. Identify the level of significance alpha and the degrees of​ freedom, d.f. = n-1. Find the critical​ value(s) using the given​ t-distribution table in the row with the correct degrees of freedom. If the hypothesis test is​ left-tailed, use the "One tail, alpha " column with a negative sign. If the hypothesis test is​ right-tailed, use the "One tail, alpha " column with a positive sign. If the hypothesis test is​ two-tailed, use the "Two tails, alpha " column with a negative and a positive sign.

Use a chi squared​-test to test the claim sigma^2 = 0.56 at the alpha = 0.10 significance level using sample statistics s^2 = 0.198 and n = 17. Assume the population is normally distributed.

Identify the test statistic. x^2 = ((n-1)*s^2)/(sigma^2) = 5.66 X^2 L = 7.96 X^2 R = 26.30 ^Although either a​ chi-square distribution table or technology can be used to find the critical​ value, for this​ explanation, a​ chi-square distribution table is used. Use a chi squared​-distribution table to find the critical values with alpha = 0.10 and 16 degrees of​ freedom. Since this is a​ two-tailed test, there will be two critical values.

Explain how to decide when a normal distribution can be used to approximate a binomial distribution.

If np>=5 and nq>=s​5, the normal distribution can be used.

Use the​ P-value to decide whether to reject Upper H 0 when the level of significance is​

If the​ P-value is less than or equal to the level of​ significance, then reject Upper H 0. If the​ P-value is greater than the level of​ significance, then fail to reject Upper H 0.

For the following​ information, determine whether a normal sampling distribution can be​ used, where p is the population​ proportion, alpha is the level of​ significance, p^ is the sample​ proportion, and n is the sample size. If it can be​ used, test the claim. ​Claim: p >= 0.46​; alpha = 0.06. Sample​ statistics: p^ = 0.40​, n = 140

Let q = 1-p and let q^ = 1 - p^. A normal sampling distribution can be used​ here, since np >= 5 and nq >= 5 ^n = 140 p = 0.46 q = 0.54 If a normal sampling distribution can be​ used, identify the hypotheses for testing the claim. H 0​: p >= 0.46​, Upper H a​: p < 0.46 ^A​ left-tailed test is being​ performed, since the alternative hypothesis Upper H a contains the ​< symbol ​(less than​). If a normal sampling distribution can be​ used, identify the critical​ value(s) for this test. z0 = -1.55 ^https://www.calculators.org/math/z-critical-value.php If a normal sampling distribution can be​ used, identify the rejection​ region(s). The rejection region is z < -1.55. If a normal sampling distribution can be​ used, identify standardized test statistic z. z = -1.42 ​ If a normal sampling distribution can be​ used, decide whether to reject or fail to reject the null hypothesis and interpret the decision. Fail to reject the null hypothesis. There is not enough evidence to reject the claim.

A security expert claims that exactly 13​% of all homeowners have a home security alarm. State Upper H 0 and Upper H Subscript a in words and in symbols. Then determine whether the hypothesis test for this claim is​ left-tailed, right-tailed, or​ two-tailed. Explain your reasoning.

State the null hypothesis in words and in symbols. Choose the correct answer below. D. The null hypothesis expressed in words​ is, "the proportion of all homeowners who own a home security alarm is 0.13​." The null hypothesis is expressed symbolically​ as, ​"Upper H0​: p = 0.13​." State the alternative hypothesis in words and in symbols. Choose the correct answer below. The alternative hypothesis expressed in words​ is, "the proportion of all homeowners who own a home security alarm is not equal to 0.13​." The alternative hypothesis is expressed symbolically​ as, ​"Upper H a​: p =/= 0.13​." The hypothesis test is two - tailed because the alternative hypothesis contains =/=.

Find the critical​ value(s) for a​ left-tailed z-test with alpha = 0.02. Include a graph with your answer.

The critical​ value(s) is(are) - 2.05. excel: =NORMSINV(0.02)

Use a​ t-test to test the claim about the population mean mu at the given level of significance alpha using the given sample statistics. Assume the population is normally distributed. ​Claim: mu = 53,000​; alpha = 0.05 Sample​ statistics: x bar = 53,990​, s=1600​, n=21

The standardized test statistic t is given by the formula​ below, where x bar is the sample​ mean, mu is the hypothesized​ mean, s is the sample standard​ deviation, and n is the sample size. The standardized test statistic is 2.84 Determine the critical​ values, rounding to three decimal places. t 0 = -2.086​, 2.086 If there is nothing unlikely or surprising about the observed parameter​ value, the magnitude of the test statistic is small. A test statistic with a high magnitude indicates that the observed parameter value would be very unlikely if the null hypothesis were true. For a​ two-tailed test, if the test​ statistic, t = 2.84​, is < than the negative critical​ value,- 2.086​, or > the positive critical​ value, 2.086​, reject the null hypothesis.​ Otherwise, fail to reject the null hypothesis.

Reject

There is enough evidence to reject the claim

Fail to reject

There is not enough evidence to support the researcher's claim

A random sample of 80 eighth grade​ students' scores on a national mathematics assessment test has a mean score of 288. This test result prompts a state school administrator to declare that the mean score for the​ state's eighth graders on this exam is more than 285. Assume that the population standard deviation is 38. At alpha = 0.04​, is there enough evidence to support the​ administrator's claim? Complete parts​ (a) through​ (e).

Write the claim mathematically and identify Upper H 0 and Upper H Subscript a. Choose the correct answer below. Upper H 0​: mu <= 285 Upper H a​: mu > 285 ​(claim) ^ A null hypothesis Upper H 0 is a statistical hypothesis that contains a statement of​ equality, such as <=​, =​, or >=. The alternative hypothesis Upper H a is the complement of the null hypothesis. It is a statement that must be true if the null hypothesis is​ false, and it contains a statement of strict​ inequality, such as >​, =/=​, or <. The claim is what you believe. ​(b) Find the standardized test statistic​ z, and its corresponding area. z = 0.71 ^The standardized test statistic is z= (x bar - mu) / (sigma / SqRoot n) where x bar is the sample​ mean, mu is the hypothesized mean​ value, sigma is the population standard​ deviation, and n is the sample size. ​(c) Find the​ P-value. ​P-value = 0.239 ​ ^Since Upper H a contains the > inequality symbol, this is a​ right-tailed test. In a​ right-tailed test the​ P-value is the area to the right of the test statistic. Subtract 0.761 from 1 to find the area to the right of z = 0.71. =1-(NORM.DIST(0.71,0,1,TRUE)) ^If test is​ two-tailed, the​ P-value is equal to twice the area in the tail of the standardized test statistic. If z is​ negative, the tail is on the left. (d) Decide whether to reject or fail to reject the null hypothesis. Fail to reject Upper H 0 ^If the​ P-value is < the level of significance alpha​, reject the null hypothesis. If the​ P-value is > the level of significance alpha​, fail to reject the null hypothesis. (e) Interpret your decision in the context of the original claim. At the 4​% significance​ level, there is not enough evidence to support the​ administrator's claim that the mean score for the​ state's eighth graders on the exam is more than 285. ^The original claim that the mean score for the​ state's eighth graders on the exam is more than 285 is the alternative​ hypothesis, Upper H a. If we reject the null​ hypothesis, there is enough evidence to support the claim. If we fail to reject the null​ hypothesis, there is not enough evidence to support the claim.

The lengths of time​ (in years) it took a random sample of 32 former smokers to quit smoking permanently are listed. Assume the population standard deviation is 5.4 years. At alphaequals0.02​, is there enough evidence to reject the claim that the mean time it takes smokers to quit smoking permanently is 13 ​years? Complete parts​ (a) through​ (e). Sample mean: 14.0875 (32 #s)

​(a) Identify the claim and state the null hypothesis and alternative hypothesis. ​: mu = 13 ​(claim) Upper H a​: mu =/= 13 (b) Identify the standardized test statistic. Use technology. z = 1.14 (c) Find the​ P-value. Use technology. P = 0.254 ^Since the test is​ two-tailed, the​ P-value is equal to two times the area in the tail of the standardized test statistic z = 1.14. P = 0.254 (d) Decide whether to reject or fail to reject the null hypothesis and​ (e) interpret the decision in the context of the original claim at the 2​% level of significance. Fail to reject Upper H 0. There is not sufficient evidence to reject the claim that the mean time it takes smokers to quit smoking permanently is 13 years.


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