Question

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 a higher or lower proportion is appropriate.  Company B randomly samples 400 people and 88 of them prefer Company A's product (Brand A).
   Assuming a 5% significance level, which one of the following statements is correct?

   		This is a one-tailed test. Accept the null hypothesis because the test statistic value of 1.0 is less than the critical value of 1.96.
   		Do not reject the null hypothesis that Company A's claim is correct.
   		This is a two-tailed test. Reject the null hypothesis because the test statistic exceeds the critical value.
   		Reject the null hypothesis. This is a one-tail test with a test statistic 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 random sample of n = 250 customers of the club. She would now like to determine whether or not the mean age of her customers is over 30. If so, she plans to alter the entertainment to appeal to an older crowd. If not, no entertainment changes will be made.

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

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

1 points   

QUESTION 4

1. In a recent survey of 600 adults, 16.4% indicated that they had fallen asleep in front of the television in the past month.  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 statistic is 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 spent on the internet per week was 9.5 hours with a standard deviation of 2.0 hours

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

   The upper limit (to three decimal places) of a 95% confidence interval for the average amount of time spent weekly on the internet 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 average monthly balance of credit card holders is equal to $750.

   An auditor selects a random sample of 100 accounts and finds that the average owed is $830.40 with a sample standard deviation of $236.50. This sample is used to perform an appropriate hypothesis test to test whether the average balance is not $750

   At a 5% level of significance the positive critical value of this 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 to be calculated. A random sample of six observations from a normal population is to be used on which to base the estimate. If the population variance is unknown the tabulated numerical value that must be used to find the upper limit of the confidence interval is:

   		z = 2.576
   		t = 3.707
   		t = 4.032
   		z = 3.291

Type or paste question here

Answer 1

Solution(1)
Given in the question
Z is a standard normal random variable than we need to calculate
P(Z>0) = ?
From Z table we found P(Z>0) = 0.5
So its answer is D.
Solution(2)
Given in the question
Null hypothesis H0: p = 0.2
Alternate Hypothesis Ha: p 0.2
No. of sample = 400
So Sample proportion p^ = 88/400 = 0.22
Test statistic = (p^ -p)/sqrt(p*(1-p)/n) = (0.22 - 0.2)/sqrt(0.2*(1-0.2/400)) = 1
Critical value at alpha = 0.05 and this is two tailed test so test critical value is +/- 1.96
If test stat value is greater than 1 or less than -1 than reject the null hypothesis else do not reject the null hypothesis. Here we can see that test stat value is in between +/-1.96 so we are failed to reject the null hypothesis.
Do not reject the null hypothesis that the company's A's claim is correct.

Solution(3)
Given in the question
No. of sample = 250
Population mean = 30
Sample mean = 30.45
Sample standard deviation = 5
Observed test statistic = (Sample mean - Population mean)/Sample standard deviation/Sqrt(n) = (30.45-30)/5/sqrt(250) = 1.423
Solution(4)
No. of sample = 600
P(Fallen Asleep) p= 0.164
98% confidence interval can be calculated as
p +/- Zalpha/2*Sqrt(p*(1-p)/n)
here alpha = 0.02, so alpha/2 = 0.01, from Z table we found Zalpha/2 = 2.33
0.164 + /- 2.33*sqrt(0.164*(1-0.164)/600)
0.164 +/- 2.33*0.015
0.164 +/- 0.035
So 98% confidence interval is 0.129 and 0.199