QUESTION 1 If Z is a standard normal random variable, then P(Z > 0) =   0 1 0.4579 0.5 1 points    QUESTION...

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Statistics

QUESTION 1

  1. If Z is a standard normal random variable, then P(Z > 0)=  

    0

    1

    0.4579

    0.5

1 points   

QUESTION 2

  1. Company A claims that 20% of people in Sydney prefer its product(Brand A). Company B disputes the 20% but has no idea whether ahigher or lower proportion is appropriate.  Company Brandomly samples 400 people and 88 of them prefer Company A'sproduct (Brand A).
    Assuming a 5% significance level, which one of the followingstatements is correct?

    This is a one-tailed test. Accept the null hypothesis becausethe test statistic value of 1.0 is less than the critical value of1.96.

    Do not reject the null hypothesis that Company A's claim iscorrect.

    This is a two-tailed test. Reject the null hypothesis becausethe test statistic exceeds the critical value.

    Reject the null hypothesis. This is a one-tail test with a teststatistic value of 1.75 and a critical value of 1.645

1 points   

QUESTION 3

  1. The owner of a local nightclub has recently surveyed a randomsample of n = 250 customers of the club. She would now like todetermine whether or not the mean age of her customers is over 30.If so, she plans to alter the entertainment to appeal to an oldercrowd. If not, no entertainment changes will be made.

    Suppose she found that the sample mean was 30.45 years and thesample standard deviation was 5 years.

    The calculated value (to three decimal places) of the teststatistic is

1 points   

QUESTION 4

  1. In a recent survey of 600 adults, 16.4% indicated that they hadfallen asleep in front of the television in the pastmonth.  Which of the following intervals represents a 98%confidence interval?

    0.129 to 0.199

    0.117 to 0.211

    0.161 to 0.167

    0.145 to 0.183

1 points   

QUESTION 5

  1. In an upper-tail hypothesis test the calculated test statisticis z = 1.08. The p-value is closest to:

    0.1401

    0.2802

    0.3431

    0.3599

1 points   

QUESTION 6

  1. The average amount of time a random sample of 25 teenagers spenton the internet per week was 9.5 hours with a standard deviation of2.0 hours

    Assume that the distribution of the amount of time spent weeklyon the internet by teenagers is normal.

    The upper limit (to three decimal places) of a 95% confidenceinterval for the average amount of time spent weekly on theinternet by teenagers is

1 points   

QUESTION 7

  1. To locate the rejection region:

    the level of α must be specified.

    the level of β must be specified.

    both  α and β must be specified.

    neither α nor β need be specified.

1 points   

QUESTION 8

  1. A bank manager would like to determine whether the averagemonthly balance of credit card holders is equal to $750.

    An auditor selects a random sample of 100 accounts and findsthat the average owed is $830.40 with a sample standard deviationof $236.50. This sample is used to perform an appropriatehypothesis test to test whether the average balance is not $750

    At a 5% level of significance the positive critical value ofthis hypothesis test is

1 points   

QUESTION 9

  1. If a random sample of size n is drawn from a normal population,then the sampling distribution of the sample mean will be:

    normal for all values of n.

    normal only for n > 30.

    approximately normal for all values of n.

    approximately normal only for n > 30.

1 points   

QUESTION 10

  1. The 99% confidence interval estimate of a population mean is tobe calculated. A random sample of six observations from a normalpopulation is to be used on which to base the estimate. If thepopulation variance is unknown the tabulated numerical value thatmust be used to find the upper limit of the confidence intervalis:

    z = 2.576

    t = 3.707

    t = 4.032

    z = 3.291

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Answer & Explanation Solved by verified expert
3.8 Ratings (497 Votes)
Solution1Given in the questionZ is a standard normal random variable than we need tocalculatePZ0 From Z table we found PZ0 05So its answer is DSolution2Given in the questionNull hypothesis H0 p 02Alternate Hypothesis Ha p 02No of    See Answer
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