BNAD test 2 (conceptual)

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Mu is < or equal to 25 and mu is > 25 ... n=40 and sample mean= 26.4 ... the population standard deviation is 6 1. test statistic 2. p value 3. at alpha of .05 what is your conclusion

1. 1.48 2. 0.0694 3. dont reject

105 blue at .24 72 brown at .13 89 green at .20 84 orange at .16 70 red at .13 80 yellow at .14 alpha of .05 test stat and critical value

5.85 11.07

The degrees of freedom for a contingency table with 10 rows and 11 columns is

90

Two approaches to drawing a conclusion in a hypothesis test are

p-value and critical value

The chi-square test of a contingency table is a test of independence for:

two qualitative variables

A goodness of fit test is always conducted as a

upper tail test

matched sample design?

when the same person is participating in two experiments

For a sample size of 30, changing from using the standard normal distribution to using the t distribution in a hypothesis test,

will result in the rejection region being smaller

sample of 244 homeowners of the 100 that fell behind before law was enacted 30 avoided forclosure of the 144 fell behind after law was enacted 48 avoide forclosure identify pvalue

.2915

n1=35 mean=13.6 s=5.2 n2=40 mean=10.1 s=8.5 test stat degrees of freedom conclusion at .05

2.18 65 reject

test statistic is -2.25 what is the p value

.0122

Read the t statistic from the table of t distributions and circle the correct answer. A two-tailed test, a sample of 20 at a .20 level of significance; t =

1.328

Read the t statistic from the table of t distributions and circle the correct answer. A one-tailed test (upper tail), a sample size of 18 at a .05 level of significance t =

1.740

In order to test if the mean returns on a major index have changed from the historic monthly average of 1.2%, a sample of 36 recent monthly returns is used to calculate the value of the relevant t-test statistic. At the 5% level of significance, we reject the null hypothesis if this value is .

Greater than 2.03 or less than -2.03

A statistical test conducted to determine whether to reject or not reject a hypothesized probability distribution for a population is known as a

goodness of fit test

A one-tailed test is a

hypothesis test in which rejection region is in one tail of the sampling distribution

An example of statistical inference is

hypothesis testing

To construct an interval estimate for the difference between the means of two populations when the standard deviations of the two populations are unknown but equal, we must use a t distribution with (let n 1 be the size of sample 1 and n 2 the size of sample 2)

(n 1 + n 2 - 2) degrees

Read the t statistic from the table of t distributions and circle the correct answer. A one-tailed test (lower tail), a sample size of 10 at a .10 level of significance; t =

-1.383

Read the z statistics from the normal distribution table and circle the correct answer. A two-tailed test at a .0694 level of significance; z =

-1.48 and 1.48

Read the z statistic from the normal distribution table and circle the correct answer. A one-tailed test (lower tail) at a .063 level of significance; z =

-1.53

48% of drivers did not stop at a stop sign... p=.48 and p does not =.48 alpha= .05 n=800 360 did not stop test statistic? does it differ from .48?

-1.7 no, p value is greater then alpha

a local courier service advertises that its average delivery time is less than 6 hours for local deliveries. In order to test this claim, a random sample of 16 of the couriers deliveries produced a sample mean time of 5.8 hours with a sample standard deviation of .25 hours u>= 6 u<6 critical values at 1% sign. conclusion

-2.602 reject

mu = 18 mu does not = 18... n=48 xbar=14 s=4.32 test stat pvalue conclusion at .05

-6.42 -1.678 reject |-6.42|> 1.678

now: occupied rooms= 1470 total rooms= 1750 prior: occupied rooms= 1458 total rooms= 1800 alpha=.05 hypothesis test if there has been an increase point estimate of proportion of rooms occupied test stat p value

.84 and .81 test stat is 2.33 pvalue .0099 reject

n1=40 mean= 25.2 pop SD= 5.2 n2= 50 mean 2= 22.8 pop SD= 6 test stat pvalue conclusion at .05

2.03 .0212 reject

a travel magazine conducts annual survey of readers favorite cruise ship. Scores based on 100 point scale. A sample of 37 ships that carry fewer then 500 passengers returned an average rating of 85.36 and a sample of 40 ships that carried 500 or more passengers had an average rating of 81.4. Population SD for ships with fewer then 500 passengers is 4.55 and for ships with 500 or more is 3.97 point estimate difference margin of error at 95% conclusion

3.96 1.91 [2.05, 5.817] reject

A one-tailed hypothesis test of the population mean has

Only one critical value

When conducting a hypothesis test for a given sample size, if the probability of a Type I error decreases, then the .

Probability of incorrectly accepting the null increases

A car dealer who sells only late-model luxury cars recently hired a new salesman and believes that this salesman is selling at lower markups. He knows that the long-run average markup in his lot is $5600. He takes a random sample of 16 of the new salesman&#39;s sales and finds an average markup of $5000 and a standard deviation of $800. Assume the markups are normally distributed. What is the value of an appropriate test statistic for the car dealer to use to test his claim?

T15=-3

In a one-tailed test, the rejection region is located under one tail (left or right) of the corresponding probability distribution, while in a two-tailed test this region is located under both tails. (TRUE OR FALSE)

TRUE

xbar=1000 xbar=1016 s=23 s=24 n=12 n=15 construct a 95% CI

[-34.8012, 2.8012]

In hypothesis testing, the critical value is

a number that establishes the boundary of the rejection region

what type of test should be performed to compare the mean SAT scores of males with the mean SAT scores of females? given sample data, it is necessary to compute the sample SD

a t test under independent sampling when the population variances are unknown

A Type II error is committed when

a true alternative hypothesis is mistakenly rejected

An important application of the chi-square distribution is

a. making inferences about a single population variance b. testing for goodness of fit c. testing for the independence of two variables ALL OF THE ABOVE

If we are interested in testing whether the proportion of items in population 1 is larger than the proportion of items in population 2, the

alternative hypothesis should state P 1 - P 2 > 0

30. If two independent large samples are taken from two populations, the sampling distribution of the difference between the two sample means

approximated by normal distribution

In order not to violate the requirements necessary to use the chi-square distribution, each expected frequency in a goodness of fit test must be

at least 5

For a two-tailed test with a sample size of 40, the null hypothesis will not be rejected at a 5% level of significance if the test statistic is

between -1.96 and 1.96, exclusively

A company wants to identify which of two production methods has the smaller completion time. One sample of workers is selected and each worker first uses one method and then uses the other method. The sampling procedure being used to collect completion time data is based on

matched samples

The level of significance is the

maximum allowable probability of Type I error

When testing the difference between two population means under independent sampling, we use the z distribution if .

population variances are known

A p-value is the

probability, when the null hypothesis is true, of obtaining a sample result that is at least as unlikely as what is observed

why do we round down for degrees of freedom for a t value

provides a better estimate and a larger t value

The level of significance in hypothesis testing is the probability of

rejecting a true null hypothesis

A two-tailed test is performed at a 5% level of significance. The p-value is determined to be 0.09. The null hypothesis

should not be rejected

2008.. 46% gave presents... in 2009 only 35% say they will.... n=60 p value conclusion

test stat= -1.71 p value= .0436 reject

For a multinomial experiment, which of the following is true?

the trials are independent

a tutor promises to improve SAT scores of students by more then 50 points after three lessons. To see if this is true the tutor takes a sample of 49 students scores after and before. the mean difference was 53 points better after tutoring, with a standard deviation of the difference equal to 12 points, Let Ud denote the mean of the difference value of test statistic critical value with alpha of .05 conclusion

z-1.75 t48=1.677 reject


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