Exam 8: Inferences Based on a Single

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Solve the problem. -A rejection region is established in each tail of the sampling distribution for a two-tailed test.

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Consider the following printout. HYPOTHESIS: MEAN X=xX = x X= gpa X = \text { gpa } SAMPLE MEAN OF X=2.9528\mathrm { X } = 2.9528 SAMPLE VARIANCE OF X=0.226933X = 0.226933 SAMPLE SIZE OF X=167X = 167 HYPOTHESIZED VALUE (x)=3( \mathrm { x } ) = 3 MEAN X - x =-0.0472 z =-1.2804 Suppose a two-tailed test is desired. Find the pp -value for the test.

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For the given binomial sample size and null-hypothesized value of P0,\mathrm { P } 0 , determine whether the sample size is large enough to use the normal approximation methodology to conduct a test of the null hypothesis H0:p"P0\mathrm { H } _ { 0 } : \mathrm { p } " \mathrm { P } 0 -The business college computing center wants to determine the proportion of business students who have laptop computers. If the proportion exceeds 30%30 \% , then the lab will scale back a proposed enlargement of its facilities. Suppose 250 business students were randomly sampled and 75 have laptops. Find the rejection region for the corresponding test using α=.05\alpha = .05 .

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In a test of H0:μ=12H _ { 0 } : \mu = 12 against Ha:μ>12H _ { \mathrm { a } } : \mu > 12 , a sample of n=75n = 75 observations possessed mean xˉ\bar { x } =13.1= 13.1 and standard deviation s=4.3s = 4.3 . Find and interpret the pp -value for the test.

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In a test of hypothesis, the sampling distribution of the test statistic is calculated under the assumption that the alternative hypothesis is true.

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A method currently used by doctors to screen women for possible breast cancer fails to detect cancer in 15% of women who actually have the disease. A new method has been developed that researchers hope will be able to detect cancer more accurately. A random sample of 80 women known to have breast cancer were screened using the new method. Of these, the new method failed to detect cancer in 9. Calculate the test statistic used by the researchers for the corresponding test of hypothesis.

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The smaller the p-value in a test of hypothesis, the more significant the results are.

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The hypotheses for H0:μ=125.4H _ { 0 } : \mu = 125.4 and Ha:μ125.4H _ { \mathrm { a } } : \mu \neq 125.4 are tested at α=.10\alpha = .10 . Sketch the appropriate rejection region.

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In a test of H0:μ=250H _ { 0 } : \mu = 250 against Ha:μ250H _ { \mathrm { a } } : \mu \neq 250 , a sample of n=100n = 100 observations possessed mean xˉ=247.3\bar { x } = 247.3 and standard deviation s=11.4s = 11.4 . Find and interpret the pp -value for the test.

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An educational testing service designed an achievement test so that the range in student scores would be greater than 420 points. To determine whether the objective was achieved, the testing service gave the test to a random sample of 30 students and found that the sample mean and variance were 759 and 1943, respectively. Conduct the test for H0:σ2=4900H _ { 0 } : \sigma ^ { 2 } = 4900 vs. Ha:σ2>4900H _ { \mathrm { a } } : \sigma ^ { 2 } > 4900 using α=.025\alpha = .025 . Assume the range is 6σ6 \sigma .

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We never conclude "Accept H0H _ { 0 } " in a test of hypothesis. This is because:

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Consider a test of H0:μ=70\mathrm { H } _ { 0 } : \mu = 70 performed with the computer. SPSS reports a two-tailed p-value of 0.0718. Make the appropriate conclusion for the given situation: Ha:μ<70,z=1.8,α=0.05\mathrm { H } _ { \mathrm { a } } : \mu < 70 , z = - 1.8 , \alpha = 0.05

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The State Association of Retired Teachers has recently taken flak from some of its members regarding the poor choice of the associationʹs name. The associationʹs by-laws require that more Than 60 percent of the association must approve a name change. Rather than convene a meeting, it Is first desired to use a sample to determine if meeting is necessary. Suppose the association Decided to conduct a test of hypothesis using the following null and alternative hypotheses: =0.6 :>0.6 Define a Type I Error in the context of this problem.

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How many tissues should a package of tissues contain? Researchers have determined that a person uses an average of 68 tissues during a cold. Suppose a random sample of 2500 people yielded the Following data on the number of tissues used during a cold: x = 55, s = 23. Identify the null and Alternative hypothesis for a test to determine if the mean number of tissues used during a cold is Less than 68.

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Increasing numbers of businesses are offering child-care benefits for their workers. However, one union claims that more than 90% of firms in the manufacturing sector still do not offer any child-care benefits. A random sample of 340 manufacturing firms is selected, and only 35 of them offer child-care benefits. Specify the rejection region that the union will use when testing at α=.10\alpha = .10

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For the given binomial sample size and null-hypothesized value of P0,\mathrm { P } 0 , determine whether the sample size is large enough to use the normal approximation methodology to conduct a test of the null hypothesis H0:p"P0\mathrm { H } _ { 0 } : \mathrm { p } " \mathrm { P } 0 - n=50,p0=0.8\mathrm { n } = 50 , \mathrm { p } 0 = 0.8

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A recipe submitted to a magazine by one of its subscribers states that the mean baking time for a cheesecake is 55 minutes. A test kitchen preparing the recipe before it is published in the magazine makes the cheesecake 10 times at different times of the day in different ovens. The following baking times (in minutes)are observed. 54 55 58 59 59 60 61 61 62 65 Assume that the baking times belong to a normal population. Test the null hypothesis that the mean baking time is 55 minutes against the alternative hypothesis μ>55\mu > 55 . Use α=.05\alpha = .05 .

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A random sample of n=15n = 15 observations is selected from a normal population to test H0:μ=2.89H _ { 0 } : \mu = 2.89 against Ha:μ<2.89H _ { \mathrm { a } } : \mu < 2.89 at α=.01\alpha = .01 . Specify the rejection region.

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α=0.1, p-value =0.005\alpha = 0.1 \text {, p-value } = 0.005

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A random sample of nn observations, selected from a normal population, is used to test the null hypothesis H0:σ2=155H _ { 0 } : \sigma ^ { 2 } = 155 . Specify the appropriate rejection region. Ha:σ2155,n=10,α=.05H _ { \mathrm { a } } : \sigma ^ { 2 } \neq 155 , n = 10 , \alpha = .05

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